Capacitance (Class 12): what happens with the battery connected or removed
The classic question: you pull the plates of a charged capacitor apart or slide a dielectric in. Does the charge, the voltage or the energy change? The answer depends only on one fact: is the battery still connected? Try both and compare.
NCERT Class 12 Physics, Chapter 2 (Electrostatic Potential and Capacitance). Every lab below runs in 3D in your browser or on a phone, in English and Hindi.
Try the labs
Parallel-plate capacitor: battery connected or disconnected (changing d and inserting a dielectric) · free · Chapter 2
To see how the capacitance, charge, potential difference, field and stored energy of a parallel-plate capacitor change when the distance between the plates is changed or a dielectric is inserted…
Connecting two capacitors: charge sharing and energy loss · plan · Chapter 2
To study charge sharing between two capacitors: the common potential and the energy lost (as heat in the resistance).
Capacitors in series/parallel, charging energy · plan · Chapter 2
To find the equivalent capacitance of series and parallel combinations of capacitors and the charge, potential difference and energy on each.
Capacitor with a dielectric slab · plan · Chapter 2
To study the effect of a dielectric slab on the capacitance, charge, voltage, field and energy of a parallel plate capacitor.
Equipotential surfaces and potential energy · plan · Chapter 2
To understand electric potential and equipotential surfaces, the relation between E and V, and potential energy of a charge.
Parallel plate capacitor with dielectric · plan · Chapter 2
To study the capacitance of a parallel plate capacitor, the effect of a dielectric slab, energy stored and the RC charging process.
Van de Graaff generator · plan · Chapter 2
To understand how a Van de Graaff generator builds up a very high potential and what limits it.
The basics
A parallel plate capacitor of plate area A and separation d, filled with a dielectric of constant K, has capacitance C = K ε₀ A / d. It holds charge Q = C V, the field between the plates is E = V / d, and the stored energy is U = ½ C V² = Q² / 2C.
C = K ε₀ A / dQ = C VU = ½ C V² = Q² / 2CE = V / d
Battery connected: V stays constant
The battery holds the voltage fixed. If you increase d, C falls, so the charge Q = C V falls too (charge flows back into the battery), the field E = V / d falls and the energy ½ C V² falls. Inserting a dielectric raises C by K, so Q and U become K times larger.
Battery disconnected: Q stays constant
With no path for charge, Q is trapped. Increasing d lowers C, so V = Q / C rises, but E = Q / (ε₀ A) stays the same. The energy Q² / 2C rises, and the extra energy is the work you do pulling the plates apart. Inserting a dielectric makes V and U fall to 1/K of their old values, and E falls to 1/K as well.
Frequently asked questions
What happens to the energy of a capacitor when the plates are pulled apart?
With the battery connected, the energy falls because C falls at constant V. With the battery removed, it rises because C falls at constant Q, and the increase equals the work done.
Does the electric field change when d changes with the battery off?
No. With Q fixed, E = Q / (ε₀ A) does not depend on d.
What does a dielectric do to the capacitance?
It multiplies the capacitance by the dielectric constant K, because the dielectric reduces the field and so the voltage for the same charge.
Why does charge flow when the battery stays connected?
The battery keeps V fixed. When C changes, Q = C V must change, so charge moves to or from the plates through the battery.
What is the energy stored in a capacitor?
U = ½ C V² = ½ Q V = Q² / 2C.