Capacitor With Battery vs Battery Removed: What Changes
Physics · Electrostatic Potential And Capacitance · NEET
The one rule that decides everything: if the battery stays connected, the voltage V is held constant; if the battery is removed (capacitor isolated), the charge Q is trapped and stays constant. So when you insert a dielectric or move the plates, first ask "what is fixed?" then use Q = CV. Memory hook: "Battery ON, Volts frozen; Battery OFF, Charge locked."
Same dielectric slab, two setups. Left: battery connected holds V constant, so C, Q and energy all rise by K while the field stays the same. Right: battery removed traps Q, so V, energy and field all fall by K. Always fix the constant quantity first.
Your doubts, answered
If the battery stays connected and I insert a dielectric, does the charge change?
Yes. With the battery connected, V is fixed. Inserting a dielectric of constant K raises capacitance to C' = KC. Since Q = CV and V is fixed, the charge rises to Q' = KCV = KQ. The battery pushes in extra charge to keep V the same. Energy U = (1/2)CV^2 also rises by K.
When the battery is removed, what stays constant, charge or voltage?
Charge stays constant. Once you disconnect the battery, there is no path for charge to leave or enter, so Q is trapped. Now if you insert a dielectric, C' = KC, and since Q is fixed, V' = Q/C' = V/K (voltage falls). Energy U = Q^2/(2C) falls by K.
Does the capacitance C itself depend on whether the battery is there?
No. C depends only on geometry (C = epsilon0 A / d) and the medium (dielectric constant K). The battery does not change C. The battery only decides which quantity, V or Q, is held constant while C changes. This is the most common exam confusion.
Battery connected: I move the plates closer. What happens to charge and energy?
Battery connected means V is constant. Moving plates closer means d decreases, so C = epsilon0 A/d increases. Then Q = CV increases (battery supplies more charge), and U = (1/2)CV^2 increases. This is exactly the NEET 2024 question.
Why does energy DECREASE when a dielectric is inserted in an isolated (battery removed) capacitor?
Q is fixed, so U = Q^2/(2C). Inserting the dielectric raises C to KC, so U falls to U/K. The lost energy is used to pull the dielectric in (the capacitor sucks the slab inward), so the field does work on the slab. Energy is not destroyed, it converts to work.
⚠️ The NEET trap ✗ Assuming the same quantity stays constant in both cases, or thinking capacitance C changes because of the battery. ✓ Battery CONNECTED means V is constant (use Q = CV, U = (1/2)CV^2). Battery REMOVED means Q is constant (use V = Q/C, U = Q^2/(2C)). C changes only due to geometry or dielectric, never because of the battery itself. 🧠 Before any dielectric or plate step, write down the ONE fixed quantity first. Battery there = V fixed. Battery gone = Q fixed.
Real NEET questions
NEET 2024
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then: 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. Choose the most appropriate answer.
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 is connected, so V is constant. Plates moved closer means d decreases. Step 1: C = epsilon0 A / d, so C increases (statement C true). Step 2: Q = CV with V fixed, so Q increases (statement A true). Step 3: U = (1/2)CV^2 with V fixed and C up, so U increases, meaning B ('decreases') is false. Step 4: QV = CV^2 increases (statement E true). Step 5: Q/V = C, and C changed, so D ('remains same') is false. Correct statements: A, C, E, which is option (a).
NEET 2022
A capacitor of capacitance C = 900 pF is charged fully by a 100 V battery B. Then B is disconnected and connected to another uncharged capacitor of capacitance C = 900 pF. The electrostatic energy stored by the system is:
A · 4.5 x 10^-6 J
B · 3.25 x 10^-6 J
C · 2.25 x 10^-6 J ✓
D · 1.5 x 10^-6 J
Solution: Battery is disconnected, so the charge is fixed. Step 1: Initial energy U_i = (1/2)CV^2 = (1/2)(900 x 10^-12)(100)^2 = 4.5 x 10^-6 J. Step 2: The total charge Q = CV = 900 x 10^-12 x 100 = 9 x 10^-8 C is now shared over two equal capacitors, so total capacitance becomes 2C and common voltage becomes V/2 = 50 V. Step 3: Final energy U_f = (1/2)(2C)(V/2)^2 = (1/4)CV^2 = U_i / 2 = 2.25 x 10^-6 J, option (c). Half the energy is lost as heat in the connecting wires.
Solved Electrostatic Potential And Capacitance NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the single rule to solve every 'with vs without battery' problem?
Ask what is held constant. Battery connected: V is constant, so use Q = CV and U = (1/2)CV^2. Battery removed: Q is constant, so use V = Q/C and U = Q^2/(2C). Then apply the change in C (from dielectric or plate movement).
When a dielectric is inserted, what happens in each case?
C always becomes KC. Battery connected (V fixed): Q becomes KQ (up), E field stays the same (E = V/d), U becomes KU (up). Battery removed (Q fixed): V becomes V/K (down), E becomes E/K (down), U becomes U/K (down).
Does the electric field between the plates depend on the battery?
With the battery connected, V is fixed so E = V/d stays constant (for fixed d). With the battery removed, Q is fixed so the field depends on charge; inserting a dielectric reduces the net field to E/K because bound charges partly cancel it.
Why is energy lost when a charged capacitor is connected to an uncharged one?
After the battery is removed, charge is shared and the common voltage drops. The final energy (1/2)(2C)(V/2)^2 is half the initial (1/2)CV^2. The missing half is dissipated as heat in the wires and as electromagnetic radiation, even with ideal wires.
Is capacitance C affected by voltage or charge?
No. C = epsilon0 A / d for a parallel plate capacitor, so it depends only on plate area, separation and the medium. It does not depend on Q or V. The battery only fixes which of Q or V stays constant while C changes.
