NCERT Solutions · Class 12 Physics Electrostatic Potential and Capacitance

By the superposition principle, the net potential at a point on the line joining the charges is the algebraic sum \(\displaystyle V=\dfrac{1}{4\pi\epsilon_0}\left(\dfrac{q_1}{r_1}+\dfrac{q_2}{r_2}\right)\), with \(\displaystyle q_1=5\times10^{-8}\,\mathrm{C}\), \(\displaystyle q_2=-3\times10^{-8}\,\mathrm{C}\), separation \(\displaystyle 16\,\mathrm{cm}\), using \(\displaystyle k=9\times10^{9}\,\mathrm{N\,m^2/C^2}\).Case $\displaystyle 1$ — between the charges. Let the null point be at distance \(\displaystyle x\) from the \(\displaystyle 5\times10^{-8}\,\mathrm{C}\) charge (so \(\displaystyle 16-x\) from the other). Setting \(\displaystyle V=0\): \[\frac{5\times10^{-8}}{x}=\frac{3\times10^{-8}}{16-x}\ \Rightarrow\ 5(16-x)=3x\ \Rightarrow\ 80=8x\ \Rightarrow\ x=10\,\mathrm{cm}\]Case $\displaystyle 2$ — outside the charges, beyond the smaller (negative) charge. No null point exists beyond the larger charge; on the far side of the \(\displaystyle -3\times10^{-8}\,\mathrm{C}\) charge, let \(\displaystyle x\) be measured from the \(\displaystyle 5\times10^{-8}\,\mathrm{C}\) charge: \[\frac{5\times10^{-8}}{x}=\frac{3\times10^{-8}}{x-16}\ \Rightarrow\ 5(x-16)=3x\ \Rightarrow\ 2x=80\ \Rightarrow\ x=40\,\mathrm{cm}\]The potential is zero at $\displaystyle 10$ cm from the \(\displaystyle 5\times10^{-8}\,\mathrm{C}\) charge (between the two charges), and at $\displaystyle 40$ cm from the \(\displaystyle 5\times10^{-8}\,\mathrm{C}\) charge, i.e. $\displaystyle 24$ cm beyond the \(\displaystyle -3\times10^{-8}\,\mathrm{C}\) charge (outside, on the far side of the negative charge).

For a regular hexagon of side \(\displaystyle a\), the distance from the centre to every vertex equals \(\displaystyle a\) itself (the vertices lie on a circle of radius \(\displaystyle a\)). Here \(\displaystyle a=10\,\mathrm{cm}=0.1\,\mathrm{m}\), and each vertex carries \(\displaystyle q=5\,\mu\mathrm{C}=5\times10^{-6}\,\mathrm{C}\).Potential is a scalar, so the potential at the centre is just the sum of the six individual contributions, each at the same distance \(\displaystyle a\): \[V=6\times\frac{kq}{a}=6\times\frac{(9\times10^{9})(5\times10^{-6})}{0.1}\] \[V=6\times4.5\times10^{5}=2.7\times10^{6}\,\mathrm{V}\]\(\displaystyle V = 2.7\times10^{6}\,\mathrm{V}\)

(a) For two equal and opposite charges, the plane that perpendicularly bisects the line AB (i.e. passes through the midpoint of AB, at right angles to AB) is an equipotential surface — every point on it is equidistant from \(\displaystyle +2\,\mu\mathrm{C}\) and \(\displaystyle -2\,\mu\mathrm{C}\), so the two contributions to \(\displaystyle V\) cancel and \(\displaystyle V=0\) everywhere on this plane.(b) The electric field is always perpendicular to an equipotential surface. Since this equipotential plane is perpendicular to AB, the field at every point on it is directed along AB, from A (the \(\displaystyle +2\,\mu\mathrm{C}\) charge) toward B (the \(\displaystyle -2\,\mu\mathrm{C}\) charge), i.e. normal to the plane.

By Gauss's law, treating the uniformly charged conducting sphere's field (outside) as that of a point charge \(\displaystyle q\) at the centre, with \(\displaystyle R=12\,\mathrm{cm}=0.12\,\mathrm{m}\) and \(\displaystyle q=1.6\times10^{-7}\,\mathrm{C}\):(a) Inside the sphere: all the charge resides on the (conducting) surface, so the enclosed charge for any Gaussian surface inside is zero. \[E_{\text{inside}}=0\](b) Just outside the sphere (\(\displaystyle r=R=0.12\,\mathrm{m}\)): \[E=\frac{kq}{R^2}=\frac{(9\times10^{9})(1.6\times10^{-7})}{(0.12)^2}=\frac{1440}{0.0144}=1.0\times10^{5}\,\mathrm{N/C}\] (directed radially outward)(c) At \(\displaystyle r=18\,\mathrm{cm}=0.18\,\mathrm{m}\): \[E=\frac{kq}{r^2}=\frac{(9\times10^{9})(1.6\times10^{-7})}{(0.18)^2}=\frac{1440}{0.0324}\approx4.44\times10^{4}\,\mathrm{N/C}\] (directed radially outward)

For a parallel-plate capacitor, \(\displaystyle C=\dfrac{K\epsilon_0 A}{d}\), so \(\displaystyle C\) is directly proportional to \(\displaystyle K\) and inversely proportional to \(\displaystyle d\). Originally (air, \(\displaystyle K=1\)): \(\displaystyle C_0=\dfrac{\epsilon_0 A}{d_0}=8\,\mathrm{pF}\).Now \(\displaystyle d_{\text{new}}=d_0/2\) and \(\displaystyle K=6\): \[C=\frac{K\epsilon_0 A}{d_0/2}=2K\cdot\frac{\epsilon_0 A}{d_0}=2K\,C_0=2\times6\times8\,\mathrm{pF}\]\(\displaystyle C=96\,\mathrm{pF}\)

(a) For capacitors in series, reciprocals add. With each \(\displaystyle C=9\,\mathrm{pF}\): \[\frac{1}{C_{\text{eq}}}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}\ \Rightarrow\ C_{\text{eq}}=3\,\mathrm{pF}\](b) In a series combination, the same charge \(\displaystyle Q\) flows onto every capacitor: \[Q=C_{\text{eq}}V=(3\,\mathrm{pF})(120\,\mathrm{V})=360\,\mathrm{pC}\] Since all three capacitors are equal, the $\displaystyle 120$ V supply divides equally among them: \[V_{\text{each}}=\frac{Q}{9\,\mathrm{pF}}=\frac{360\,\mathrm{pC}}{9\,\mathrm{pF}}=40\,\mathrm{V}\]Each capacitor has a potential difference of $\displaystyle 40$ V across it.

(a) For capacitors in parallel, capacitances simply add: \[C_{\text{eq}}=2+3+4=9\,\mathrm{pF}\](b) In a parallel combination, each capacitor has the full supply voltage \(\displaystyle V=100\,\mathrm{V}\) across it, so \(\displaystyle Q=CV\) for each individually: \[Q_1=(2\,\mathrm{pF})(100\,\mathrm{V})=200\,\mathrm{pC}=2\times10^{-10}\,\mathrm{C}\] \[Q_2=(3\,\mathrm{pF})(100\,\mathrm{V})=300\,\mathrm{pC}=3\times10^{-10}\,\mathrm{C}\] \[Q_3=(4\,\mathrm{pF})(100\,\mathrm{V})=400\,\mathrm{pC}=4\times10^{-10}\,\mathrm{C}\]

For a parallel-plate capacitor with air between the plates, \(\displaystyle C=\dfrac{\epsilon_0 A}{d}\), with \(\displaystyle A=6\times10^{-3}\,\mathrm{m}^2\), \(\displaystyle d=3\,\mathrm{mm}=3\times10^{-3}\,\mathrm{m}\), \(\displaystyle \epsilon_0=8.854\times10^{-12}\,\mathrm{C^2/N\,m^2}\): \[C=\frac{(8.854\times10^{-12})(6\times10^{-3})}{3\times10^{-3}}=1.77\times10^{-11}\,\mathrm{F}\approx17.7\,\mathrm{pF}\]With \(\displaystyle V=100\,\mathrm{V}\): \[Q=CV=(1.77\times10^{-11}\,\mathrm{F})(100\,\mathrm{V})=1.77\times10^{-9}\,\mathrm{C}\]\(\displaystyle C\approx17.7\,\mathrm{pF}\); charge on each plate \(\displaystyle \approx1.77\times10^{-9}\,\mathrm{C}\) (equal and opposite).

The mica sheet (\(\displaystyle K=6\)) is exactly as thick as the original $\displaystyle 3$ mm gap, so it fills the whole gap and the new capacitance is \(\displaystyle C'=KC_0=6\times17.7\,\mathrm{pF}\approx106\,\mathrm{pF}\) (using \(\displaystyle C_0\approx17.7\,\mathrm{pF}\) from Q2.8).(a) Voltage supply still connected — \(\displaystyle V\) is held fixed at $\displaystyle 100$ V by the battery, so \[Q'=C'V=(1.06\times10^{-10}\,\mathrm{F})(100\,\mathrm{V})\approx1.06\times10^{-8}\,\mathrm{C}\] The charge increases (from \(\displaystyle 1.77\times10^{-9}\,\mathrm{C}\) to \(\displaystyle \approx1.06\times10^{-8}\,\mathrm{C}\)) as the battery supplies more charge; since \(\displaystyle U=\tfrac12 CV^2\) with \(\displaystyle V\) fixed, the stored energy also increases by the same factor of 6.(b) Supply disconnected first — the charge \(\displaystyle Q_0\approx1.77\times10^{-9}\,\mathrm{C}\) is now isolated and stays constant. As \(\displaystyle C\) rises to \(\displaystyle C'=6C_0\), the voltage falls: \[V'=\frac{Q_0}{C'}=\frac{V_0}{K}=\frac{100\,\mathrm{V}}{6}\approx16.7\,\mathrm{V}\] Since \(\displaystyle U=\dfrac{Q_0^2}{2C}\) with \(\displaystyle Q_0\) fixed and \(\displaystyle C\) increased $\displaystyle 6$-fold, the stored energy decreases to \(\displaystyle 1/6\) of its original value.

Energy stored in a charged capacitor: \(\displaystyle U=\dfrac12 CV^2\), with \(\displaystyle C=12\,\mathrm{pF}=12\times10^{-12}\,\mathrm{F}\), \(\displaystyle V=50\,\mathrm{V}\). \[U=\frac12(12\times10^{-12})(50)^2=\frac12(12\times10^{-12})(2500)\] \[U=1.5\times10^{-8}\,\mathrm{J}\]\(\displaystyle U=1.5\times10^{-8}\,\mathrm{J}\)

Before connection, \(\displaystyle C_1=600\,\mathrm{pF}\) is charged to \(\displaystyle V_0=200\,\mathrm{V}\): \[Q_0=C_1V_0=(600\times10^{-12})(200)=1.2\times10^{-7}\,\mathrm{C}\] \[U_i=\tfrac12 C_1V_0^2=\tfrac12(600\times10^{-12})(200)^2=1.2\times10^{-8}\,\mathrm{J}\]After connection to the uncharged \(\displaystyle C_2=600\,\mathrm{pF}\), charge redistributes (charge is conserved) until both share a common potential \(\displaystyle V'\), by conservation of charge \(\displaystyle Q_0=(C_1+C_2)V'\): \[V'=\frac{Q_0}{C_1+C_2}=\frac{1.2\times10^{-7}}{1200\times10^{-12}}=100\,\mathrm{V}\] \[U_f=\tfrac12(C_1+C_2)V'^2=\tfrac12(1200\times10^{-12})(100)^2=6\times10^{-9}\,\mathrm{J}\]Energy lost (dissipated as heat/radiation during the redistribution, since this is not a reversible process): \[\Delta U=U_i-U_f=1.2\times10^{-8}-6\times10^{-9}=6\times10^{-9}\,\mathrm{J}\]\(\displaystyle \Delta U=6\times10^{-9}\,\mathrm{J}\) is lost.