NCERT Solutions Class 12 Physics Chapter – 2

Q.1 Two charges 5 × 10⁻⁸ C and −3 × 10⁻⁸ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

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$\begin{array}{l}\mathrm{Here},{\mathrm{q}}_1=5\times {10}^{-8}\mathrm{C}\\ {\mathrm{q}}_2=3\times {10}^{-8}\mathrm{C}\\ \text{Distance between the two charges},\text{d}=\text{16 cm}\\ =0.\text{16 m}\\ \text{Let\hspace{0.17em}potential\hspace{0.17em}be\hspace{0.17em}zero\hspace{0.17em}}\mathrm{at}\text{\hspace{0.17em}a\hspace{0.17em}distance\hspace{0.17em}s\hspace{0.17em}cm\hspace{0.17em}from\hspace{0.17em}charge\hspace{0.17em}}{\mathrm{q}}_1\\ =5\times {10}^{-8}\mathrm{C}\\ \therefore {\mathrm{r}}_1=\mathrm{s}\times {10}^{-2}\mathrm{m};\\ {\mathrm{r}}_2=(16-\mathrm{s})\times {10}^{-2}\mathrm{m}\\ \text{Let\hspace{0.17em}}\mathrm{electric}\text{\hspace{0.17em}potential\hspace{0.17em}at\hspace{0.17em}distance\hspace{0.17em}s}=0\\ \mathrm{Electric}\mathrm{potential}\mathrm{at}\mathrm{a}\mathrm{point}\mathrm{is}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{potentials}\\ \mathrm{caused}\mathrm{by}\mathrm{charges}{\mathrm{q}}_1\mathrm{and}{\mathrm{q}}_2\mathrm{respectively}.\\ \\ \therefore \mathrm{V}=\frac{{\mathrm{q}}_1}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}+\frac{{\mathrm{q}}_2}{4{\mathrm{\pi \epsilon }}_0(\mathrm{d}-\mathrm{r})}\to \left(\mathrm{i}\right)\\ \text{Where},{\mathrm{\epsilon }}_0=\text{Permittivity of free space}\\ \text{For V}=0,\text{equation}\left(\text{i}\right)\text{becomes}\\ 0=\frac{{\mathrm{q}}_1}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}+\frac{{\mathrm{q}}_2}{4{\mathrm{\pi \epsilon }}_0(\mathrm{d}-\mathrm{r})}\\ \frac{{\mathrm{q}}_1}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}=-\frac{{\mathrm{q}}_2}{4{\mathrm{\pi \epsilon }}_0(\mathrm{d}-\mathrm{r})}\\ \frac{{\mathrm{q}}_1}{\mathrm{r}}=-\frac{{\mathrm{q}}_2}{(\mathrm{d}-\mathrm{r})}\\ \frac{5\times {10}^{-8}}{\mathrm{r}}=-\frac{(-3\times {10}^{-8})}{(0.16-\mathrm{r})}\\ \frac{0.16}{\mathrm{r}}-\frac{0.16}{\mathrm{r}}=\frac{8}{5}\\ \therefore \mathrm{r}=0.1\mathrm{m}\\ =10\mathrm{cm}\\ \therefore \text{The potential is zero at a distance of 1}0\text{cm from the}\\ {\text{positive charge q}}_1\\ \text{If the point lies outside the charges,}\\ \left\{\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{°}}}\right\}[\frac{5\times {10}^{-8}}{\mathrm{r}}-\frac{3\times {10}^{-8}}{(\mathrm{r}-0.16)}]=0\\ \Rightarrow \mathrm{r}=40\mathrm{cm}\\ \therefore \mathrm{}\mathrm{r}\text{= 40 cm from positive charge towards negative}\\ \text{charge on extended line.}\end{array}$

Q.2 A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.

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\begin{array}{l}\text{Consider a regular hexagon}\text{ABCDEF, having}\text{an}\text{equal}\text{amount}\text{of}\text{charge}\text{q}\\ \text{at}\text{each}\text{of}\text{its}\text{vertices}\text{.}\\ {\text{Here, charge, q = 5 μC = 5×10}}^{\text{-6}}\text{C}\\ \text{Side of hexagon, l = 10 cm}\\ \text{Distance of each vertex from the}\text{centre O, d = 10 cm = 0}\text{.1 m}\\ \text{As}\text{electric}\text{potential}\text{is}\text{scalar,}\\ \therefore \text{Electric potential at O}\text{is}\text{given}\text{as}\\ \text{V =}\frac{\text{6q}}{\text{4}\pi {\text{ε}}_{\text{0}}\text{d}}\\ \text{Here,}{\text{ε}}_{\text{0}}\text{= Permittivity of free space}\\ \frac{\text{1}}{\text{4}\pi {\text{ε}}_{\text{0}}}{\text{= 9×10}}^{\text{9}}{\text{NC}}^{\text{-2}}{\text{m}}^{\text{-2}}\\ \therefore \text{V =}\frac{{\text{6×9×10}}^{\text{9}}{\text{NC}}^{\text{-2}}{\text{m}}^{\text{-2}}{\text{×5×10}}^{\text{-6}}\text{C}}{\text{0}\text{.1 m}}\text{= 2}{\text{.7×10}}^{\text{6}}\text{V}\\ \therefore \text{The potential at the centre of the hexagon is 2}{\text{.7×10}}^{\text{6}}\text{V}\text{.}\end{array}

Q.3 Two charges 2 μC and −2 μC are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?

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(a) The plane normal to AB and passing through its middle point has zero potential everywhere and it represents an equipotential surface of the given system.
(b) The direction of the electric field at every point on this surface is along normal to the plane in the direction of AB.

Q.4 A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ C distributed uniformly on its surface. What is the electric field
(a) Inside the sphere
(b) Just outside the sphere
(c) At a point 18 cm from the centre of the sphere?

Ans-

$\begin{array}{l}\left(\text{a}\right)\text{Here,\hspace{0.17em}Radius of the spherical conductor},\text{r}=\text{12 cm}\\ =0.\text{12 m}\\ \text{Charge over the\hspace{0.17em}spherical conductor},\text{q}=\text{1}.\text{6}\times \text{1}{0}^{-\text{7}}\text{C}\\ \text{If there is field inside the conductor},\text{then charges}\\ \text{will move to neutralize it}.\\ \therefore \text{Electric field inside the spherical conductor =0}\\ \left(\text{b}\right)\text{Electric field just outside the spherical\hspace{0.17em}conductor is}\\ \text{given}\mathrm{as}:\\ \mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\\ \text{Here},{\mathrm{\epsilon }}_0=\text{Permittivity of free space}\\ \frac{1}{4{\mathrm{\pi \epsilon }}_0}=9\times {10}^9{\mathrm{Nm}}^2{\mathrm{C}}^{-2}\\ \therefore \mathrm{E}=9\times {10}^9\times \frac{\text{1}.\text{6}\times \text{1}{0}^{-\text{7}}}{{(0.\text{12})}^2}\\ ={10}^5{\mathrm{NC}}^{-1}\\ \therefore \text{The electric field just outside the sphere is}{10}^5{\mathrm{NC}}^{-1}.\\ \mathrm{}\left(\text{c}\right)\text{Let\hspace{0.17em}Electric field at a point 18 m from the centre of}\\ \text{the sphere}={\text{E}}_{\text{1}}\\ \text{Distance of the\hspace{0.17em}point from the centre\hspace{0.17em}of\hspace{0.17em}sphere},\text{d}\\ =\text{18 cm}\\ =0.\text{18 m}\mathrm{}\\ {\text{E}}_{\text{1}}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\\ =9\times {10}^9\times \frac{\text{1}.\text{6}\times \text{1}{0}^{-\text{7}}}{{(0.18)}^2}\\ =4.4\times {10}^4{\mathrm{NC}}^{-1}\\ \therefore \text{The electric field at a point 18 cm from the centre of}\\ \text{the sphere is\hspace{0.17em}}4.4\times {10}^4{\mathrm{NC}}^{-1}.\end{array}$

Q.5 A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10⁻¹² F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

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$\begin{array}{l}\mathrm{As}\mathrm{initially}\mathrm{air}\mathrm{was}\mathrm{filled}\mathrm{betwen}\mathrm{the}\mathrm{plates}\mathrm{of}\mathrm{parallel}\\ \mathrm{plate}\mathrm{capacitor},\\ \mathrm{Dielectric}\mathrm{constant}\mathrm{of}\mathrm{air},\mathrm{K}=1\\ \mathrm{Initial}\mathrm{capacitance}\mathrm{of}\mathrm{the}\mathrm{parallel}\mathrm{plate}\mathrm{capacitor}\\ =8\mathrm{pF}\\ \mathrm{Let}\mathrm{area}\mathrm{of}\mathrm{each}\mathrm{plate}\mathrm{of}\mathrm{capacitor}=\mathrm{A}\\ \mathrm{Let}\mathrm{initial}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{plates}\mathrm{of}\mathrm{the}\mathrm{parallel}\\ \mathrm{plate}\mathrm{capacitor}=\mathrm{d}\\ \mathrm{Capacitance}\mathrm{of}\mathrm{the}\mathrm{parallel}\mathrm{plate}\mathrm{capacitor}\mathrm{is}\mathrm{given}\mathrm{as}:\\ {\mathrm{C}}_1=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}\\ =8\mathrm{pF}\to \left(\mathrm{i}\right)\\ \mathrm{When}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{plates}\mathrm{is}\mathrm{reduced}\mathrm{to}\mathrm{half},\\ \mathrm{then}\mathrm{new}\mathrm{distance},\mathrm{d}‘=\frac{\mathrm{d}}{2}\\ \mathrm{Dielectric}\mathrm{constant}\mathrm{of}\mathrm{substance}\mathrm{filled}\mathrm{between}\mathrm{the}\\ \mathrm{plates},\mathrm{K}‘=6\\ \\ \therefore \mathrm{Capacitance}\mathrm{of}\mathrm{the}\mathrm{parallel}\mathrm{plate}\mathrm{capacitor}\mathrm{becomes},\\ {\mathrm{C}}_2=\mathrm{K}‘\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}‘}\\ =\frac{6{\mathrm{\epsilon }}_0\mathrm{A}}{\frac{\mathrm{d}}{2}}\\ =\frac{6\times 2\times {\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}\\ =12\times \frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}\\ =12\times 8\left[\mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right)\right]\\ =96\mathrm{pF}\\ \therefore \mathrm{The}\mathrm{capacitance}\mathrm{between}\mathrm{the}\mathrm{plates}\mathrm{is}96\mathrm{pF}.\end{array}$

Q.6 Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

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$\begin{array}{l}\left(\text{a}\right)\text{Capacitance of each capacitor},\text{C}=\text{9 pF}\\ \mathrm{Since}\text{\hspace{0.17em}the\hspace{0.17em}three\hspace{0.17em}capacitors\hspace{0.17em}are\hspace{0.17em}connected\hspace{0.17em}in\hspace{0.17em}series,}\\ \therefore \text{Equivalent capacitance of the\hspace{0.17em}combination of}\\ \text{capacitors is given by the relation},\\ \frac{1}{{\text{C}}_{\text{s}}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}+\frac{1}{{\mathrm{C}}_3}\\ \frac{1}{{\text{C}}_{\text{s}}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}\\ =\frac{1}{3}\\ \therefore {\text{C}}_{\text{s}}=3\mathrm{pF}\\ \therefore \text{Total capacitance of the combination is}3\mathrm{pF}.\\ \mathrm{}\left(\text{b}\right)\text{\hspace{0.17em}Here,\hspace{0.17em}}\mathrm{s}\text{upply voltage},\text{V}=\text{12}0\text{V}\\ \text{Since\hspace{0.17em}the\hspace{0.17em}}\mathrm{three}\text{\hspace{0.17em}capacitors\hspace{0.17em}are\hspace{0.17em}connected\hspace{0.17em}in\hspace{0.17em}series,}\\ \therefore \mathrm{The}\mathrm{p}\text{otential difference}(\text{V}’)\text{across each capacitor is}\\ \text{equal\hspace{0.17em}to one}-\text{third of the supply voltage}.\\ \therefore \text{V}’=\frac{\mathrm{V}}{3}\\ =\frac{\text{12}0}{3}\\ =40\mathrm{V}\\ \therefore \text{The potential difference across each capacitor is}\\ \text{4}0\text{V}.\end{array}$

Q.7 Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

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$\begin{array}{l}\left(\text{a}\right)\mathrm{Given},{\mathrm{C}}_1=2\mathrm{pF}\\ {\mathrm{C}}_2=3\mathrm{pF}\\ {\mathrm{C}}_3=4\mathrm{pF}\\ \text{For the parallel combination of the capacitors},\text{the\hspace{0.17em}}\\ {\text{total\hspace{0.17em}capacitance\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}combination,\hspace{0.17em}C}}_{\text{p}}={\mathrm{C}}_1+{\mathrm{C}}_2+{\mathrm{C}}_3\\ =2+3+4=9\mathrm{pF}\\ \therefore \text{Total capacitance of the combination is 9 pF}.\\ \left(\text{b}\right)\text{Here,\hspace{0.17em}}\mathrm{s}\text{upply voltage},\text{V}=\text{1}00\text{V}\\ \text{The voltage through\hspace{0.17em}each\hspace{0.17em}of the three capacitors is}\\ \text{same, V}=\text{1}00\text{V}\\ \text{Charge on the capacitor of capacitance C and potential}\\ \text{difference V is given by the\hspace{0.17em}relation},\\ \text{q}=\text{VC}\dots \left(\text{i}\right)\\ \text{For C}=\text{2 pF},\\ \mathrm{Charge},{\mathrm{q}}_1=\text{VC}\\ \text{=100}\times \text{2}\\ \text{=200\hspace{0.17em}pF}\\ \text{=2}\times {10}^{-10}\mathrm{C}\\ \\ \text{For C}=\text{3 pF},\\ \mathrm{Charge},{\mathrm{q}}_2=\text{VC}\\ \text{=100}\times 3\\ \text{=300\hspace{0.17em}pF}\\ \text{=3}\times {10}^{-10}\mathrm{C}\\ \text{For C}=\text{4 pF},\\ \mathrm{Charge},{\mathrm{q}}_3=\text{VC}\\ \text{=100}\times \text{2}\\ \text{=200\hspace{0.17em}pF}\\ \text{=2}\times {10}^{-10}\mathrm{C}\end{array}$

Q.8 In a parallel plate capacitor with air between the plates, each plate has an area of 6 ×10⁻³ m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

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$\begin{array}{l}\text{Here,\hspace{0.17em}}\mathrm{a}\text{rea of each plate of the parallel plate capacitor},\\ \text{A}=\text{6}\times \text{1}{0}^{-\text{3}}{\text{\hspace{0.17em}m}}^{\text{2}}\\ \text{Separation between the plates},\text{d}=\text{3 mm}\\ =\text{3}\times \text{1}{0}^{-\text{3}}\text{m}\\ \text{Supply voltage},\text{V}=\text{1}00\text{V}\\ \text{Capacitance of a parallel plate capacitor is given by\hspace{0.17em}}\\ \text{the\hspace{0.17em}relation,}\\ \text{C=}\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}\\ \text{Here},{\mathrm{\epsilon }}_0=\text{Permittivity of free space}\\ =\text{8}.\text{854}\times \text{1}{0}^{-\text{12}}{\text{N}}^{-\text{1}}{\text{m}}^{-\text{2}}{\text{C}}^{-\text{2}}\\ \therefore \text{C=}\frac{\text{8}.\text{854}\times \text{1}{0}^{-\text{12}}\times \text{6}\times \text{1}{0}^{-\text{3}}}{\text{3}\times \text{1}{0}^{-\text{3}}}\\ =17.71\times \text{1}{0}^{-\text{12}}\mathrm{F}\\ =17.71\mathrm{pF}\\ \mathrm{Potential}\mathrm{is}\mathrm{related}\mathrm{to}\mathrm{the}\mathrm{charge}\mathrm{q}\mathrm{and}\mathrm{capacitance}\mathrm{C}\\ \mathrm{as}:\mathrm{V}=\frac{\mathrm{q}}{\mathrm{C}}\\ \therefore \mathrm{q}=\mathrm{VC}\\ =\text{1}00\times 17.71\times \text{1}{0}^{-\text{12}}\\ =1.771\times \text{1}{0}^{-9}\mathrm{C}\\ \therefore \text{Capacitance of the capacitor is 17}.\text{71 pF and charge}\\ \text{on each plate is 1}.\text{771}\times \text{1}{0}^{-\text{9}}\text{C}.\end{array}$

Q.9 Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) While the voltage supply remained connected.
(b) After the supply was disconnected.

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\begin{array}{l}\left(\text{a}\right)\text{While}\text{the}\text{voltage}\text{supply}\text{remained}\text{connected,}\\ \text{potential across the plates remains 1}00\text{V}.\\ \text{Here,}d\text{ielectric constant of the mica sheet},\text{k}=\text{6}\\ \text{Initial capacitance},\text{C}=\text{1}.\text{771}\times \text{1}{0}^{-\text{11}}\text{F}\\ \text{New}\text{capacitance}\text{is}\text{given}\text{by,}\\ \text{C’=kC}\\ \text{C’=6}\times \text{1}.\text{771}\times \text{1}{0}^{-\text{11}}=106pF\\ \text{Supply voltage},\text{V}=\text{1}00\text{V}\\ \text{New}\text{charge}\text{is}\text{given}by\text{,}\\ \text{q’=C’V}\\ =10\text{6}\times \text{1}{0}^{-\text{12}}\times 100\\ =1.06\times \text{1}{0}^{-8}\\ \left(\text{b}\right)\text{After}\text{the}\text{supply}\text{was}\text{disconnected,}theamountof\\ \text{charge remains}\text{constant}\text{.}\\ \text{Here,}d\text{ielectric constant},\text{k}=\text{6}\\ \text{Initial capacitance},\text{C}=\text{1}.\text{771}\times \text{1}{0}^{-\text{11}}\text{F}\\ \text{New}\text{capacitance}\text{is}\text{given}\text{as:}\text{C’=kC}\\ \text{=6}\times \text{1}.\text{771}\times \text{1}{0}^{-\text{11}}\\ =106pF\\ \text{Charge,}\text{q}=\text{1}.\text{771}\times \text{1}{0}^{-\text{9}}\text{C}\\ \text{Potential across the plates is given by}\text{the}\text{relation:}\\ \text{V’=}\frac{q}{C‘}\\ =\frac{\text{1}.\text{771}\times \text{1}{0}^{-\text{9}}}{106\times \text{1}{0}^{-\text{12}}}\\ =16.7V\end{array}

Q.10 A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

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\begin{array}{l}\text{Here,}c\text{apacitor of the capacitance},\text{C}=\text{12 pF}\\ =\text{12}\times \text{1}{0}^{-\text{12}}\text{F}\\ \text{Potential difference},\text{V}=\text{5}0\text{V}\\ \text{Electrostatic energy stored in the capacitor is given as:}\\ \text{E=}\frac{1}{2}C{V}^2\\ =\frac{1}{2}\times \text{12}\times \text{1}{0}^{-\text{12}}\times {\left(50\right)}^2\\ =1.5\times {10}^{-8}J\\ \therefore \text{The electrostatic energy stored in the capacitor is}\\ 1.5\times {10}^{-8}J.\end{array}

Q.11 A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

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\begin{array}{l}\text{Here,}c\text{apacitance of the capacitor},\text{C}=\text{6}00\text{pF}\\ Supplyvoltage,\text{V}=\text{2}00\text{V}\\ \text{Electrostatic energy stored in the capacitor is given by}\\ \text{the}\text{relation},\\ E=\frac{1}{2}C{V}^2\\ =\frac{\left(600\times {10}^{-12}\right)\times {\left(\text{2}00\right)}^2}{2}\\ =1.2\times {10}^{-5}J\\ \text{When}\text{supply voltage}\text{is disconnected from the}\\ \text{capacitor and another capacitor of capacitance C}=\\ \text{6}00\text{pF is connected to it},\text{then equivalent capacitance}\\ \left(\text{C}’\right)\text{of the combination is given}as:\\ \\ \frac{1}{C‘}=\frac{1}{C}+\frac{1}{C}\\ =\frac{1}{600}+\frac{1}{600}\\ =\frac{1}{300}\\ \therefore \text{C}’=300\text{pF}\\ \text{New electrostatic energy can be obtained as:}\\ \text{E’=}\frac{1}{2}\times \text{C}’\times V^2\\ =\frac{1}{2}\times 300\times {\left(200\right)}^2\\ =0.6\times {10}^{-5}J\\ Lossofel\text{ectrostatic energy}=E-E‘\\ =1.2\times {10}^{-5}-0.6\times {10}^{-5}\\ =0.6\times {10}^{-5}\\ =6\times {10}^{-6}J\\ \therefore \text{The electrostatic energy lost in the process is}\\ 6\times {10}^{-6}J.\end{array}

Q.12 A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10⁻⁹ C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).

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\begin{array}{l}\text{Here,}c\text{harge located at origin},\text{q}=\text{8 mC}\\ =\text{8}\times \text{1}{0}^{-\text{3}}\text{C}\\ \text{Magnitude of the charge},\text{which is to}\text{be}\text{carried from}\\ \text{point P to point Q},{\text{q}}_{\text{1}}=-\text{2}\times \text{1}{0}^{-\text{9}}\text{C}\\ \text{All the points are shown in the given figure}.\\ \text{Distance}\text{of}p\text{oint P},{\text{d}}_{\text{1}}=\text{3 cm},\\ \text{from the origin along z}-\text{axis}.\\ \text{Distance}\text{of}p\text{oint Q},{\text{d}}_{\text{2}}=\text{4 cm},\\ \text{from the origin along y}-\text{axis}.\\ \text{Potential at point P}\text{is}\text{given}\text{as:}{V}_1=\frac{q}{4\pi {\epsilon }_0\times d_1}\\ \text{Potential at point Q}\text{is}\text{given}\text{as:}{V}_2=\frac{q}{4\pi {\epsilon }_0\times d_2}\\ \text{Since}w\text{ork done}\left(\text{W}\right)\text{by the electrostatic force is}\\ \text{independent of the path,}\\ \therefore W=q_1\left[V_2-V_1\right]\\ \\ =q_1\left[\frac{q}{4\pi {\epsilon }_0\times d_2}-\frac{q}{4\pi {\epsilon }_0\times d_1}\right]\\ =\frac{q_{1}q}{4\pi {\epsilon }_0}\left[\frac{1}{d_2}-\frac{1}{d_1}\right]\to (i)\\ Here,\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^{-2}\\ \therefore W=9\times {10}^9\times \text{8}\times \text{1}{0}^{-\text{3}}\times \left(-\text{2}\times \text{1}{0}^{-\text{9}}\right)\left[\frac{1}{0.04}-\frac{1}{0.03}\right]\\ =-144\times \text{1}{0}^{-3}\times \left(\frac{-25}{3}\right)\\ =1.27J\\ \therefore W\text{ork done during the process is 1}.\text{27 J}.\end{array}

Q.13 A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

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\begin{array}{l}\text{Here, side of the cube}=\text{b}\\ \text{Charge at each}\text{vertex}\text{of}the\text{cube}=\text{q}\\ \text{A cube of side b is shown in the given figure}.\\ \text{Here,}\text{d}=\text{Diagonal}\text{of}the\text{cube}\text{of}each\text{side}\text{b}\\ d=\sqrt{b^2+b^2}\\ =\sqrt{2b^2}\\ =\sqrt{2}b\\ \therefore d=b\sqrt{2}\\ \text{l}=\text{Length of the diagonal of the cube}\\ l=\sqrt{d^2+b^2}\\ =\sqrt{{\left(\sqrt{2}b\right)}^2+b^2}\\ =\sqrt{2b^2+b^2}\\ =\sqrt{3b^2}=\sqrt{3}b\\ \therefore Dis\tan cebetweenthecentreofcubeandeach\\ vertex,r=\frac{l}{2}\\ =\frac{\sqrt{3}b}{2}\\ \text{Electric potential}\left(\text{V}\right)\text{at the centre of the cube is}\\ \text{because}\text{of the presence of eight}\text{charges at the}\\ \text{vertices}.\\ \therefore V=\frac{8q}{4\pi {\epsilon }_{0}r}\\ =\frac{8q}{4\pi {\epsilon }_0\left(\frac{\sqrt{3}b}{2}\right)}\\ =\frac{4q}{\sqrt{3}\pi {\epsilon }_{0}b}\\ \therefore \text{The potential at the centre of the cube is}\frac{4q}{\sqrt{3}\pi {\epsilon }_{0}b}.\\ \text{The electric field at the centre of the cube},\text{due to the}\\ \text{eight charges},\text{gets cancelled, because the charges are}\\ \text{distributed symmetrically w}\text{.r}\text{.t}\text{. the centre of the}\text{cube}.\\ \therefore \text{The electric field is zero at the centre}.\end{array}

Q.14 Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Ans-

\begin{array}{l}\text{Two charges placed at points A and B are represented}\\ \text{in the given figure}.\text{O is the midpointof the line joining}\\ \text{the two charges}.\\ \text{Amount of charge located at A},{\text{q}}_{\text{1}}=\text{1}\text{.5}\mu \text{C}\\ \text{Amount of charge located at B},{\text{q}}_{\text{2}}=\text{2}.\text{5}\mu \text{C}\\ \text{Distance between the charges},\text{d}=\text{3}0\text{cm}=0.\text{3 m}\\ \left(\text{a}\right){\text{Let V}}_{\text{1}}{\text{and E}}_{\text{1}}\text{be the electric potential and electric}\\ \text{field respectively at O}.\\ {\text{V}}_{\text{1}}=\text{Potential due to charge at A}+\text{Potential due to}\\ \text{charge at B}\\ {\text{V}}_{\text{1}}=\frac{{\text{q}}_{\text{1}}}{4\pi {\epsilon }_0\left(\frac{d}{2}\right)}+\frac{{\text{q}}_2}{4\pi {\epsilon }_0\left(\frac{d}{2}\right)}\\ =\frac{1}{4\pi {\epsilon }_0\left(\frac{d}{2}\right)}\left({\text{q}}_{\text{1}}+{\text{q}}_2\right)\\ \text{Here},{\epsilon }_0=\text{Permittivity of free space}\\ \frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{C}^2{m}^{-2}\\ \therefore {\text{V}}_{\text{1}}=\frac{9\times {10}^9\times {10}^{-6}}{\left(\frac{0.30}{2}\right)}\left(2.5+1.5\right)\\ =2.4\times {10}^{5}V\\ {\text{E}}_{\text{1}}=\text{Electric field due to}{\text{charge q}}_{\text{2}}-\text{Electric field}\\ \text{due to charge}{\text{q}}_{\text{1}}\\ =\frac{{\text{q}}_2}{4\pi {\epsilon }_0{\left(\frac{d}{2}\right)}^2}-\frac{{\text{q}}_1}{4\pi {\epsilon }_0{\left(\frac{d}{2}\right)}^2}\\ =\frac{9\times {10}^9}{{\left(\frac{0.30}{2}\right)}^2}\times {10}^6\times \left(2.5-1.5\right)\\ =4\times {10}^{5}V{m}^{-1}\\ \therefore A\text{t the}\text{mid}-\text{point},t\text{he potential is 2}.\text{4}\times \text{1}{0}^{\text{5}}\text{V}\\ \text{and electric field is}\text{4}\times \text{1}{0}^{\text{5}}{\text{V m}}^{-\text{1}}.\\ \text{The direction}\text{of}\text{the}electric\text{field is from the larger}\\ \text{charge to the smaller charge}.\end{array}

\begin{array}{l}\left(\text{b}\right)\text{Consider a point Z such that the}\text{normal distance}\\ \text{OZ}=\text{1}0\text{cm}=0.\text{1 m},\text{as shown in the}given\\ \text{figure}.\\ {\text{V}}_{\text{2}}{\text{and E}}_{\text{2}}\text{respectively}\text{are}\text{the electric potential and}\\ \text{electric field at Z}.\\ \text{From the figure,}i\text{t can be observed that distance},\\ BZ=AZ\\ =\sqrt{{\left(0.1\right)}^2+{\left(0.15\right)}^2}\\ =0.18m\\ {\text{V}}_{\text{2}}=\text{Electric potential due to A}+\text{Electric Potential due}\\ \text{to B}\\ \text{=}\frac{q_1}{4\pi {\epsilon }_0\left(AZ\right)}+\frac{q_1}{4\pi {\epsilon }_0\left(BZ\right)}\\ =\frac{9\times {10}^9\times {10}^{-6}}{0.18}\left(1.5+2.5\right)\\ =2\times {10}^{5}V\\ \\ \text{Electric field due to q at Z}\text{is}\text{given}\text{by},\\ E_A=\frac{q_1}{4\pi {\epsilon }_0{\left(AZ\right)}^2}\\ =\frac{9\times {10}^9\times 1.5\times {10}^{-6}}{{\left(0.18\right)}^2}\\ =0.416\times {10}^{6}V{m}^{-1}\\ E=\sqrt{E_A{}^2+E_B{}^2+2E_A{E}_B\cos 2\theta }\\ \text{Here},\text{2}\theta \text{is the angle},\angle AZB\\ \text{From the figure},\text{we get}\\ \cos \theta =\frac{0.10}{0.18}\\ =\frac{5}{9}\\ =0.5556\\ \therefore \theta ={\cos }^{-1}0.5556\\ =56.25\\ \therefore 2\theta ={112.5}^o\\ \cos 2\theta =-0.38\\ \text{E=}\sqrt{{\left(\text{0}{\text{.416×10}}^{\text{6}}\right)}^{\text{2}}\text{×}{\left(\text{0}{\text{.69×10}}^{\text{6}}\right)}^{\text{2}}\text{+2×0}\text{.416×0}{\text{.69×10}}^{\text{12}}\text{×}\left(\text{-0}\text{.38}\right)}\\ =6.6\times {10}^{5}V{m}^{-1}\\ \therefore \text{The potential at a point 1}0\text{cm}\\ \left(\text{perpendicular to the mid}-\text{point}\right)\\ \text{is 2}.0\times \text{1}0\text{5 V}\text{and electric field is 6}.\text{6}\times \text{1}{0}^{\text{5}}{\text{V m}}^{-\text{1}}.\end{array}

Q.15 A spherical conducting shell of inner radius r₁ and outer radius r₂ has a charge Q.
(a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Ans-

(a) The magnitude of charge placed at the centre of the shell is +q. Therefore, a charge −q will be induced to the inner surface of the shell.

$\begin{array}{l}\therefore \text{Total charge on the inner surface of the shell =}-\text{q}.\\ \text{Surface charge density at the inner surface of the shell}\\ \text{is given\hspace{0.17em}as:}\\ {\mathrm{\sigma }}_{\text{1}}=\frac{\mathrm{Total}\mathrm{charge}}{\mathrm{Inner}\mathrm{surface}\mathrm{area}}\\ =\frac{-\mathrm{q}}{4{\mathrm{\pi r}}_1^2}\to \left(\mathrm{i}\right)\\ \text{A charge of\hspace{0.17em}magnitude\hspace{0.17em}}+\text{q induces on the outer surface}\\ \text{of the shell}.\\ \text{Since\hspace{0.17em}}\mathrm{a}\text{charge of magnitude Q is placed on the outer}\\ \text{surface\hspace{0.17em}of the shell}.\\ \therefore \text{Total charge on the outer\hspace{0.17em}surface of the shell= Q}+\text{q}\\ \text{Surface charge density at the outer surface of the}\\ \text{shell\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}}\mathrm{as}:\\ {\mathrm{\sigma }}_2=\frac{\mathrm{Total}\mathrm{charge}}{\mathrm{Outer}\mathrm{surface}\mathrm{area}}\\ =\frac{\mathrm{Q}+\mathrm{q}}{4{\mathrm{\pi r}}_2^2}\to \left(\mathrm{i}\right)\end{array}$

(b) Yes, the electric field intensity inside a cavity with no charge is zero, even if the shell is not spherical and has any irregular shape. When a closed loop is taken such that a part of it is inside the cavity along a field line and the rest is inside the conductor; the net work done by the electric field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field inside a cavity with no charge is zero, whatever is the shape.

\begin{array}{l}\text{(a) Show that the normal component of electrostatic field has a discontinuity}\\ \text{from one side of a charged surface to another given by}\end{array}

\begin{array}{l}\\ \left({\overrightarrow{E}}_2-{\overrightarrow{E}}_1\right)\cdot \widehat{n}\text{=}\frac{\text{σ}}{{\text{ε}}_{\text{0}}}\\ \text{Where, (}\widehat{n})\text{is a unit vector normal to the surface at a point and σ is the surface}\\ \text{charge density at that point}\text{. (The direction of (}\widehat{n})\text{is from side 1 to side 2}\text{.)}\\ \text{Hence show that just outside a conductor, the electric field is σ}\frac{\widehat{n}}{{\text{ε}}_{\text{0}}}.\end{array}

\begin{array}{l}\\ \text{(b) Show that the tangential component of electrostatic field is continuous from}\\ \text{one side of a charged surface to another}\text{.}\\ [\text{Hint: For (a), use Gauss’s law}\text{. For, (b) use the fact that work done by electrostatic}\\ \text{field on a closed loop is zero}\text{.}]\end{array}

\begin{array}{l}\left(\text{a}\right)\text{Let}e{\text{lectric field on one side of a charged body is E}}_{\text{1}}\\ \text{and electric field on the other side of the same body is}\\ {\text{E}}_{\text{2}}.\text{If the}\text{infinite plane charged body has a uniform}\\ \text{thickness},\text{then}\\ \text{Electric field due to one surface of the charged body is}\\ \text{given as:}\\ \overrightarrow{E_1}=-\frac{\sigma }{2{\epsilon }_0}\widehat{n}\to (i)\\ \text{Here},\widehat{n}=\text{Unit vector normal to the surface at a point}\\ \sigma =\text{Surface charge density at that point}\\ \text{Electric field due to the other surface of the charged}\\ \text{body}\text{is}\text{given}\text{as:}\\ \overrightarrow{E_2}=\frac{\sigma }{2{\epsilon }_0}\widehat{n}\to (ii)\\ \text{Electric field at any point due to the two surfaces}\text{is}\\ \text{given}\text{as:}\\ \overrightarrow{E_2}-\overrightarrow{E_1}=\frac{\sigma }{2{\epsilon }_0}\widehat{n}+\frac{\sigma }{2{\epsilon }_0}\widehat{n}\\ =\frac{\sigma }{{\epsilon }_0}\widehat{n}\\ \left(\overrightarrow{E_2}-\overrightarrow{E_1}\right).\widehat{n}=\frac{\sigma }{{\epsilon }_0}\\ \text{As, inside the closed conductor},\overrightarrow{E_1}=0,\\ \\ \therefore \overrightarrow{E}=\overrightarrow{E_2}\\ =-\frac{\sigma }{2{\epsilon }_0}\widehat{n}\\ \therefore \text{Electric field just outside the conductor is}\frac{\sigma }{{\epsilon }_0}\widehat{n}.\\ \left(\text{b}\right)\text{If a charged particle is moved from one point to}\text{the}\\ \text{other on a closed loop},\\ \text{Work done by the electrostatic field=0}\\ \therefore \text{The tangential component of}\text{electrostatic field is}\\ \text{continuous from one side of a charged surface to the}\\ \text{other}.\end{array}

Q.16 A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Ans-

\begin{array}{l}\text{Charge density of the long charged cylinder of length L}\\ \text{and radius a=}\lambda .\\ \text{Another cylinder of same length surrounds the}\\ \text{previous cylinder,}\text{its radius= b}\\ \text{Let electric field produced in the space between the}\\ \text{two cylinders=E}\\ \text{According}\text{to Gauss}’\text{s theorem,}e\text{lectric flux through the}\\ \text{cylinderical}\text{Gaussian surface is given by,}\\ \varphi =\left(2\pi r\right)L\\ \text{Here},\text{r}=\text{Distance of a point from the common axis}\\ \text{of the two}\text{cylinders}\\ \text{Let total charge on the cylinder=q}\\ \text{According}\text{to}\text{Gauss’s}\text{theorem}\\ \varphi =E\left(2\pi rL\right)=\frac{q}{{\epsilon }_0}\\ \text{Here},\text{q}=\text{Charge on the inner sphere of the outer}\\ \text{cylinder}\\ \text{Since,}\text{q=}\lambda \text{L}\\ {\epsilon }_0=\text{Permittivity of free space}\\ \therefore E\left(2\pi rL\right)=\frac{\lambda \text{L}}{{\epsilon }_0}\\ \therefore E=\frac{\lambda }{2\pi {\epsilon }_{0}r}\\ \therefore \text{The electric field in the space between the two}\\ \text{cylinders is}\frac{\lambda }{2\pi {\epsilon }_{0}r}.\end{array} 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Q.17 In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?

Ans-

\begin{array}{l}\text{Here,}\text{distance between electron}\text{and}\text{proton in a}\\ \text{hydrogen atom},d=0.53\stackrel{o}{A}\\ \text{Charge of}\text{electron},{\text{q}}_{\text{1}}=-\text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\text{C}\\ \text{Charge of}\text{proton},{\text{q}}_{\text{2}}=+\text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\text{C}\\ \left(\text{a}\right)\text{The}p\text{otential at infinity is zero}.\\ \text{P}\text{.E}\text{.}\text{of}\text{the}\text{system=P}\text{.E}\text{.}at\inf inity-P.E.atdis\tan ced\\ =0-\frac{{\text{q}}_{\text{1}}{\text{q}}_{\text{2}}}{4\pi {\epsilon }_{0}d}\\ \text{Here},{\epsilon }_0\text{=Permittivity of free space}\\ \frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}N{m}^2{C}^{-2}\\ \therefore P.E.=0-\frac{9\times {10}^9\times {\left(\text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\right)}^2}{0.53\times {10}^{-10}}\\ =-43.47\times \text{1}{0}^{-\text{19}}J\\ Since1.6\times \text{1}{0}^{-\text{19}}J=1eV\\ \therefore P.E.=-43.47\times \text{1}{0}^{-\text{19}}J\\ =\frac{-43.47\times \text{1}{0}^{-\text{19}}}{1.6\times \text{1}{0}^{-\text{19}}}eV\\ =-\text{27}.\text{2 eV}\\ \therefore \text{P}\text{.E}\text{. of the system is}-\text{27}.\text{2 eV}.\\ \left(\text{b}\right)\text{As,}\text{K}\text{.E}\text{. in}\text{the}\text{orbit=}\frac{1}{2}\left(P.E.\right)\\ \therefore K.E.=\frac{1}{2}\times \left(-\text{27}.\text{2}\right)\\ =13.6eV\\ \text{Total energy}=\text{13}.\text{6}-\text{27}.\text{2}\\ =\text{13}.\text{6 eV}\\ \therefore \text{The minimum work required to free the electron is}\\ \text{13}.\text{6 eV}.\\ \left(\text{c}\right)\text{When zero of P}\text{.E}\text{. is taken},\\ d_1=1.06\stackrel{o}{A}\\ \therefore \text{P}\text{.E}\text{. of the system}=\text{P}\text{.E}{\text{. at d}}_{\text{1}}-\text{P}\text{.E}\text{. at d}\\ \text{=}\frac{q_1{q}_2}{4\pi {\epsilon }_0{d}_1}-\text{27}.\text{2}\text{eV}\\ =\frac{9\times {10}^9\times {\left(\text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\right)}^2}{1.06\times {10}^{-10}}-\text{27}.\text{2}\text{eV}\\ =21.73\times \text{1}{0}^{-\text{19}}J-\text{27}.\text{2}\text{eV}\\ =13.58eV-\text{27}.\text{2}\text{eV}\\ =-13.6eV\end{array} \begin{array}{l}{}_{}\end{array}

Q.18 Two charged conducting spheres of radii a and b are connected to each other by a wire.
What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Ans-

\begin{array}{l}\text{Let the radius of a sphere A}\text{=}\text{a}\\ \text{Let}\text{charge on}\text{the sphere}\text{A}\text{=}{\text{Q}}_{\text{A}}\\ \text{Let capacitance of the sphere}\text{A}\text{=}{\text{C}}_{\text{A}}\\ \text{Let radius of a sphere B}\text{=}\text{b},\\ \text{Let charge on}\text{sphere}\text{B}\text{=}{\text{Q}}_{\text{B}}\\ \text{Let}\text{capacitance of the sphere}\text{B}\text{=}{\text{C}}_{\text{B}}\\ As\text{the two spheres are connected with a wire},\text{their}\\ \text{potentials}\left(\text{V}\right)\text{become equal}.\\ {\text{Let electric field of sphere A=E}}_{\text{A}}\\ \text{Let}{\text{electric field of sphere B=E}}_{\text{B}}\\ \therefore R\text{atio},of{\text{E}}_{\text{A}}to{\text{E}}_{\text{B}},\\ \frac{{\text{E}}_{\text{A}}}{{\text{E}}_{\text{B}}}=\frac{{\text{Q}}_{\text{A}}}{4\pi {\epsilon }_0\times a^2}\times \frac{b^2\times 4\pi {\epsilon }_0}{{\text{Q}}_{\text{B}}}\\ \frac{{\text{E}}_{\text{A}}}{{\text{E}}_{\text{B}}}=\frac{{\text{Q}}_{\text{A}}}{{\text{Q}}_{\text{B}}}\times \frac{b^2}{a^2}\to (i)\\ But,\frac{{\text{Q}}_{\text{A}}}{{\text{Q}}_{\text{B}}}=\frac{{\text{C}}_{\text{A}}V}{{\text{C}}_{B}V}\\ and\frac{{\text{C}}_{\text{A}}}{{\text{C}}_B}=\frac{a}{b}\to (ii)\\ \text{Substituting the value of}\left(ii\right)\text{in}\left(i\right),\text{we get}\\ \frac{{\text{E}}_{\text{A}}}{{\text{E}}_{\text{B}}}=\frac{a}{b}\times \frac{b^2}{a^2}\\ =\frac{b}{a}\\ \therefore \text{The ratio of electric fields at the surface is}\frac{b}{a}.\end{array}