Mastery of electrostatic potential and capacitance is essential for Class 12 students because this chapter unifies the ideas of potential, field, energy and charge distribution — concepts that frequently appear in board examinations and competitive tests (JEE/NEET) in both conceptual and calculation-based formats. Problems involving capacitors, dielectrics, energy changes and potential gradients test a student’s ability to connect mathematical relations with physical reasoning.
Beyond routine formula use, the chapter trains students in handling multi-step scenarios: work done by fields, energy accounting when connections change, image-method reasoning for conductors, and interpretation of potential-versus-distance graphs. These abilities are repeatedly tested in higher-demand problems on the board and in competitive exams, so targeted practice on this chapter yields high returns.
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Marking
Q1. A parallel‑plate capacitor of capacitance is charged to and then disconnected from the battery. A dielectric slab is inserted completely between the plates, increasing the capacitance to . Neglecting fringe effects, the final energy stored in the capacitor is:
Q2. Assertion (A): If the electric potential at a point is , then the electric field at that point must be zero.
Reason (R): The electric field is , so depends on the spatial variation of (its gradient), not on the absolute value of .
Both (A) and (R) are true and (R) is the correct explanation of (A)
(A) is false but (R) is true
Both (A) and (R) are true but (R) is not the correct explanation of (A)
(A) is true but (R) is false
Q3. The electric potential along the ‑axis is given by
V(x)=\begin{cases}10-2x & 0\le x\le4\$$4pt] 2x-6 & 4\le x\le6\end{cases}\ \text{(volts)}.A positive test charge is moved slowly from to . The work done by the electrostatic force on the charge is:
Q4. Two isolated capacitors, (initially at ) and (initially uncharged), are disconnected from any battery and then connected in parallel (positive to positive) by ideal wires until equilibrium. The energy dissipated as heat during this redistribution is:
Q5. Assertion (A): When a dielectric slab is slowly inserted into a charged isolated (disconnected) parallel‑plate capacitor, the electrostatic energy stored in the capacitor decreases.
Reason (R): The dielectric is attracted into the capacitor by the field so the electrostatic field does positive work on the slab; that work comes from the stored electrostatic energy.
Both (A) and (R) are true but (R) is not the correct explanation of (A)
Both (A) and (R) are true and (R) is the correct explanation of (A)
(A) is true but (R) is false
(A) is false but (R) is true
Q6. Two point charges and are brought from infinity and placed apart in vacuum. The work required to bring them from infinity to this separation (electrostatic potential energy) is:
Q7. A point charge is held at a distance from an infinite grounded conducting plane. Using the method of images, the magnitude of the attractive force between the charge and the plane is:
Q8. Assertion (A): For a conductor in electrostatic equilibrium the electric potential is the same at every point inside the conductor and on its surface.
Reason (R): In electrostatic equilibrium the electric field inside a conductor is zero, hence inside the conductor.
Both (A) and (R) are true but (R) is not the correct explanation of (A)
(A) is true but (R) is false
(A) is false but (R) is true
Both (A) and (R) are true and (R) is the correct explanation of (A)
Q9. The potential of a spherically symmetric charge distribution is given by
V(r)=\begin{cases}80\ \text{V} & r\le 0.20\ \text{m}\$$4pt] \dfrac{16}{r}\ \text{V} & r\ge 0.20\ \text{m}\end{cases}(continuous at ). The surface charge density on the spherical surface at is:
Q10. Two capacitors and are charged separately so that has across it and has across it (i.e., equal and opposite charges in magnitude). Their positive plates are then connected to each other by an ideal wire and allowed to reach equilibrium. The energy dissipated in the process is: