Effect of Changing Plate Separation on a Capacitor
Physics · Electrostatic Potential And Capacitance · NEET
When you change the plate separation d of a parallel plate capacitor, the answer depends on ONE thing: is the battery still connected? If the battery stays connected, V is fixed. If the battery is removed, the charge Q is trapped and fixed. Everything else follows from C = ε₀A/d. Memory hook: "Battery ON keeps V, battery OFF keeps Q."
Left: with the battery connected, V is held fixed so increasing the gap d lowers C, Q and U. Right: with the battery removed, the trapped charge Q is fixed so increasing d lowers C but raises V and the stored energy U.
Your doubts, answered
If I pull the plates apart, does the charge go up or down?
It depends on the battery. Battery connected: V is fixed, C = ε₀A/d falls when d rises, so Q = CV falls (charge decreases, some flows back to the battery). Battery removed: Q is trapped on the plates and cannot change, so Q stays exactly the same no matter how far you pull the plates.
What quantity stays constant when the capacitor is disconnected from the battery?
The charge Q. Once the wires are removed, there is no path for charge to leave, so Q is locked. Then V = Q/C changes: if you increase d, C falls and V = Q/C rises. This is the opposite of the battery-connected case where V is locked instead.
Does the energy stored increase or decrease when I move the plates apart?
Battery removed (Q fixed): use U = Q²/2C. Increasing d lowers C, so U increases. The extra energy comes from the work YOUR hand does pulling the attracting plates apart. Battery connected (V fixed): use U = ½CV². Increasing d lowers C, so U decreases (energy is pushed back into the battery).
Why does capacitance decrease when the plates are farther apart?
C = ε₀A/d. Distance d is in the denominator, so C is inversely proportional to d. Physically, wider spacing means the same voltage produces a weaker field and holds less charge per volt. Double the gap, halve the capacitance.
Which formula for energy should I use, ½CV² or Q²/2C?
Pick the one whose constant term stays fixed. Battery connected means V is constant, so use U = ½CV². Battery removed means Q is constant, so use U = Q²/2C. Choosing the wrong one is the number one mistake students make in these questions.
⚠️ The NEET trap ✗ Assuming charge Q stays constant and energy decreases in every case, so plate movement never changes Q. ✓ Battery connected = V fixed. Moving plates closer: C = ε₀A/d rises, so Q = CV rises, U = ½CV² rises, and QV rises. Charge does change here. Only when the battery is DISCONNECTED is Q the fixed quantity. 🧠 The trap in NEET 2024: the question said the battery is CONNECTED, but many students answered as if the charge were fixed.
Real NEET questions
2024
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then choose the correct statements: A. The charge stored in it increases. B. The energy stored in it decreases. C. Its capacitance increases. D. The ratio of charge to its potential remains the same. E. The product of charge and voltage increases.
A · A, C and E only ✓
B · B, D and E only
C · A, B and C only
D · A, B and E only
Solution: Battery stays connected so V is constant. Step 1: d decreases, so C = ε₀A/d increases → statement C is TRUE. Step 2: Q = CV, and C rose while V fixed, so Q increases → A is TRUE. Step 3: U = ½CV² rises since C rose → B says 'decreases', so B is FALSE. Step 4: Q/V = C, which changed → D is FALSE. Step 5: QV, with Q up and V fixed, increases → E is TRUE. Correct statements: A, C and E. Answer: option A.
2018
The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is:
A · Proportional to the square root of the distance between the plates
B · Linearly proportional to the distance between the plates
C · Independent of the distance between the plates ✓
D · Inversely proportional to the distance between the plates
Solution: Force on one plate F = Q²/(2ε₀A). The capacitor is isolated (battery removed) so Q is fixed, and A is fixed. Notice the formula contains no d term at all. Therefore F does not depend on the plate separation. Answer: option C. (Note: the force is constant even as you pull the plates apart, which is why the work done pulling them apart, W = F·Δd, goes into increasing the stored energy.)
Solved Electrostatic Potential And Capacitance NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
When the battery is disconnected and plates moved apart, what happens to voltage?
Q is fixed. C = ε₀A/d decreases as d rises. Since V = Q/C, the voltage increases. So the potential difference across an isolated capacitor rises when you separate its plates.
Does the electric field between the plates change with separation?
For an isolated capacitor (Q fixed), E = Q/(ε₀A) has no d, so the field stays constant even as you move the plates. For a battery-connected capacitor, E = V/d changes because V is fixed while d changes.
Where does the extra energy come from when plates are pulled apart with battery off?
The plates attract each other, so pulling them apart means your hand does positive work against that attraction. That work is stored as extra electrostatic potential energy, matching the rise in U = Q²/2C.
Quick summary table for NEET: battery on vs off when d increases?
Battery ON (V fixed): C down, Q down, E = V/d down, U = ½CV² down. Battery OFF (Q fixed): C down, V = Q/C up, E constant, U = Q²/2C up. Memorise which quantity is locked and derive the rest.
