Electrostatic Potential and Capacitance for NEET UG
Electric potential is energy per unit charge, and capacitors store that energy. These ideas power camera flashes, defibrillators and touch screens.
PhysicsClass 1259 MCQs3 lessons12 flashcards3 memory tricks
Practise Electrostatic Potential and Capacitance free
What you'll learn
- Electric potential and potential energy
Calculate potential due to point charges and the work done moving a charge. - Capacitors and capacitance
Find capacitance, combine capacitors, and see the effect of a dielectric. - Energy stored in a capacitor
Calculate the energy stored and what happens when capacitors share charge.
Key formulas
| Potential of a point charge | V = kq/r |
|---|---|
| Potential energy of two charges | U = kq₁q₂/r |
| Work to move charge q | W = q(V_B − V_A) |
| Field from potential | E = −dV/dr |
| Capacitance | C = Q/V |
| Parallel-plate capacitor | C = ε₀A/d |
| With a dielectric filling the gap | C = Kε₀A/d |
| In series | 1/C = 1/C₁ + 1/C₂ + … |
Memory tricks
Capacitors in parallel add (C = C₁ + C₂); in series, 1/C = 1/C₁ + 1/C₂, the reverse of resistors.
Inserting a dielectric with the battery connected keeps V fixed, so Q and energy increase. With the battery disconnected, Q stays fixed, so V and stored energy decrease (U = Q²/2C).
W = q(V_B − V_A) = 0 when V_A = V_B. The field is always perpendicular to equipotential surfaces.
Try 3 NEET-style questions
Q1. The dielectric strength of a material is
- the maximum electric field it can withstand without electrical breakdown
- its dielectric constant
- the charge it can store
- its resistance
Show answer
Answer: (1) the maximum electric field it can withstand without electrical breakdown
Dielectric strength is the maximum field a dielectric can withstand without breaking down (losing its insulating property). For air it is about 3 × 10⁶ V/m.
Q2. A slab of thickness t (less than d) and dielectric constant K is placed between the plates of a parallel plate capacitor of area A and separation d. Its capacitance becomes
- ε₀A/(d − t + t/K)
- ε₀A/(d − t)
- Kε₀A/d
- ε₀A/(d + t − t/K)
Show answer
Answer: (1) ε₀A/(d − t + t/K)
The slab behaves like an air gap of thickness t/K. The effective separation is (d − t) + t/K, so C = ε₀A/(d − t + t/K).
Q3. A charged parallel plate capacitor is disconnected from the battery and its plate separation is doubled. The energy stored in it
- halves
- stays the same
- becomes four times
- doubles
Show answer
Answer: (4) doubles
Isolated: Q is fixed. Doubling d halves C. U = Q²/2C, so U doubles. The extra energy comes from the work done pulling the plates apart.
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