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NCERT Exemplar · Class 12 Physics Electrostatic Potential and Capacitance

33 questions

Short Answer Questions 19–23 (part 4 of 5)

  1. Exercise 19

    Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.
    NCERT’s answer
    Suppose this were not true. The potential just inside the surface would be different from that at the surface resulting in a potential gradient. This would mean that there are field lines pointing inwards or outwards from the surface. These lines cannot at the other end be again on the surface, since the surface is equipotential. Thus, this is possible only if the other end of the lines are at charges inside, contradicting the premise. Hence, the entire volume inside must be at the same potential.
    Let the closed surface be at potential \(\displaystyle V_S\), with no charge inside.\[\oint \mathbf{E}\cdot d\mathbf{A}=\frac{q_{\text{in}}}{\varepsilon_0}=0 \]NCERT_Solution_Class12_Physics_Exemplar_Ch2_Q19No field line starts or ends inside. Suppose \(\displaystyle \mathbf{E}\neq 0\) somewhere inside. Its line cannot close on itself (\(\displaystyle \oint\mathbf{E}\cdot d\mathbf{l}=0\)), so it runs from a point A on the surface to a point B on it, with \(\displaystyle \mathbf{E}\cdot d\mathbf{l}>0\) all along:\[V_A-V_B=\int_A^B \mathbf{E}\cdot d\mathbf{l}>0 \]But \(\displaystyle V_A=V_B=V_S\) \(\displaystyle \Rightarrow\) contradiction. Hence\[\mathbf{E}=0 \;\Rightarrow\; V=V_S\ \text{at every point inside} \]Answer: The enclosed volume is equipotential, at \(\displaystyle V_S\).

  2. Exercise 20

    A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.
    NCERT’s answer
    C will decrease Energy stored \(\displaystyle =\frac{1}{2} C V^2\) and hence will increase. Electric field will increase. Charge stored will remain the same. \(\displaystyle V\) will increase.
    With the battery off, no charge can leave the plates, so \(\displaystyle Q\) stays fixed. Removing the dielectric takes \(\displaystyle K\to 1\).\[C=\frac{K\varepsilon_0 A}{d}\;\to\; C'=\frac{\varepsilon_0 A}{d}=\frac{C}{K}\quad(\text{decreases}) \]\[Q'=Q\quad(\text{constant}) \]\[V'=\frac{Q}{C'}=\frac{Q}{C/K}=KV\quad(\text{increases}) \]\[E'=\frac{V'}{d}=\frac{KV}{d}=KE\quad(\text{increases}) \]\[U'=\frac{Q^2}{2C'}=\frac{KQ^2}{2C}=KU\quad(\text{increases}) \]Answer: Capacitance decreases; charge remains constant; voltage, electric field and stored energy increase (each by a factor \(\displaystyle K\)).

  3. Exercise 21

    Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity.
    NCERT’s answer
    Consider any path from the charged conductor to the uncharged conductor along the electric field. The potential will continually decrease along this path. A second path from the uncharged conductor to infinity will again continually lower the potential further. Hence this result.
    Take A positive (for negative A every inequality reverses). B is neutral, so it carries induced \(\displaystyle -q'\) on the near face and \(\displaystyle +q'\) on the far face.NCERT_Solution_Class12_Physics_Exemplar_Ch2_Q21Lines from A end on the \(\displaystyle -q'\) face; lines from the \(\displaystyle +q'\) face can only go to infinity (no other conductor). Potential falls along a field line:\[V_A-V_B=\int_A^B \mathbf{E}\cdot d\mathbf{l}>0 \]\[V_B-V_\infty=\int_B^\infty \mathbf{E}\cdot d\mathbf{l}>0 \]\[V_A>V_B>V_\infty=0 \]Answer: B lies between A and infinity in potential.

  4. Exercise 22

    Calculate potential energy of a point charge −q\displaystyle -q placed along the axis due to a charge +Q\displaystyle +Q uniformly distributed along a ring of radius R\displaystyle R. Sketch P.E. as a function of axial distance z\displaystyle z from the centre of the ring. Looking at graph, can you see what would happen if −q\displaystyle -q is displaced slightly from the centre of the ring (along the axis)?
    NCERT’s answer
    \(\displaystyle \mathrm{U}=\frac{-\mathrm{q} Q}{4 \pi \varepsilon_0 R \sqrt{1+z^2 / R^2}}\) The variation of potential energy with z is shown in the figure. The charge \(\displaystyle -q\) displaced would perform oscillations. We cannot conclude anything just by looking at the graph. NCERT_Solution_Class12_Physics_Exemplar_Ch2_Q22_ncert
    Every element of the ring is at distance \(\displaystyle \sqrt{R^2+z^2}\) from the point on the axis.\[V(z)=\frac{1}{4\pi\varepsilon_0}\,\frac{Q}{\sqrt{R^2+z^2}} \]\[U(z)=(-q)\,V(z)=-\frac{1}{4\pi\varepsilon_0}\,\frac{qQ}{\sqrt{R^2+z^2}} \]NCERT_Solution_Class12_Physics_Exemplar_Ch2_Q22\(\displaystyle U\) is symmetric in \(\displaystyle z\), lowest at the centre, and tends to zero far away:\[U(0)=-\frac{qQ}{4\pi\varepsilon_0 R},\qquad U(\infty)=0 \]For a small displacement \(\displaystyle z\ll R\), with \(\displaystyle \sqrt{R^2+z^2}\approx R\left(1+\dfrac{z^2}{2R^2}\right)\):\[U\approx-\frac{qQ}{4\pi\varepsilon_0 R}\left(1-\frac{z^2}{2R^2}\right),\qquad F=-\frac{dU}{dz}=-\frac{qQ}{4\pi\varepsilon_0 R^3}\,z \]The force points back to the centre: stable equilibrium, so \(\displaystyle -q\) oscillates along the axis about the centre (mass \(\displaystyle m\)):\[\omega=\sqrt{\frac{qQ}{4\pi\varepsilon_0\,mR^3}} \]Answer: \(\displaystyle U=-\dfrac{qQ}{4\pi\varepsilon_0\sqrt{R^2+z^2}}\); on a small displacement \(\displaystyle -q\) oscillates about the centre.

  5. Exercise 23

    Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R\displaystyle R.
    NCERT’s answer
    \(\displaystyle V=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\sqrt{R^2+z^2}}\) NCERT_Solution_Class12_Physics_Exemplar_Ch2_Q23_ncert
    Every element \(\displaystyle dq\) of the ring is at the same distance from P.NCERT_Solution_Class12_Physics_Exemplar_Ch2_Q23\[r=\sqrt{R^2+z^2} \]\[dV=\frac{1}{4\pi\varepsilon_0}\,\frac{dq}{\sqrt{R^2+z^2}} \]Potential is a scalar, so the contributions add algebraically, with no direction components:\[V=\frac{1}{4\pi\varepsilon_0}\,\frac{1}{\sqrt{R^2+z^2}}\int dq=\frac{1}{4\pi\varepsilon_0}\,\frac{Q}{\sqrt{R^2+z^2}} \]Checks:\[z=0:\ V=\frac{Q}{4\pi\varepsilon_0 R},\qquad z\gg R:\ V\approx\frac{Q}{4\pi\varepsilon_0 z} \]Answer: \(\displaystyle V=\dfrac{Q}{4\pi\varepsilon_0\sqrt{R^2+z^2}}\)