Mastery of electrostatic potential and capacitance is crucial for the CBSE board because it links field concepts to energy and work — topics that appear in numericals, experiment-based questions and viva. For competitive exams (JEE/NEET) this chapter tests multi-step reasoning: relating to via , using energy methods (), and handling dielectrics, which frequently form the core of higher-difficulty problems.
Beyond formulas, the chapter trains students to interpret potential graphs, reason about stability (minima/maxima of and ), and predict what changes when circuits are altered (battery connected vs isolated). These are precisely the kinds of reasoning and calculation skills required for both board-level numerical problems and the conceptual depth of JEE/NEET questions.
15
Minutes
10
Questions
1 / -0
Marking Scheme
Q1. The electrostatic potential along the -axis for is given by (in volts). A test charge is moved quasi‑statically from to . The change in its electrostatic potential energy equals:
Q2. Two parallel metal plates of area separated by are connected to a battery. Assuming vacuum between the plates (), the magnitude of charge on each plate is closest to:
Q3. Assertion (A): For any two points lying on the same equipotential surface, the work done by the electrostatic force in moving a test charge between them is zero. Reason (R): The electric field at every point on an equipotential surface is perpendicular to the surface.
Q4. A capacitor of capacitance is charged to and then disconnected from the battery so the charge is isolated. A dielectric slab with dielectric constant is inserted completely between the plates. The change in stored electrostatic energy (in joules) is:
Q5. Two identical point charges are fixed at and on the -axis. Consider the electrostatic potential along the -axis. At the midpoint , which of the following is correct?
Q6. The graph of potential near is symmetric and can be approximated as with (so has a local minimum at ). A small test charge is placed at and slightly displaced to . For which sign of will the electrostatic force be restorative (i.e., tend to bring the charge back toward )?
Q7. Assertion (A): For an isolated charged capacitor (battery disconnected) the insertion of a dielectric slab between the plates reduces the potential difference across the plates. Reason (R): The dielectric polarises and produces an internal electric field parallel to the external field, thereby reducing the net field.
Q8. A capacitor is connected to a battery. The energy stored in the capacitor is:
Q9. A parallel-plate capacitor has initial capacitance . It is connected to a battery of voltage and charged. A dielectric slab of is inserted so it fills exactly half the plate separation (slab thickness ). If and , the work done by the battery during insertion (in joules) is:
Q10. Two capacitors and are initially isolated. is charged to while is uncharged. Then the two capacitors are connected in parallel by joining like plates (positive-to-positive, negative-to-negative) and allowed to come to equilibrium. The heat dissipated during this process is: