Physics · Electrostatic Potential And Capacitance · NEET
Work = charge × voltage only when the voltage is constant. In a capacitor the voltage is NOT constant while charging: it starts at 0 (empty plates) and rises to V (full). At any instant the plate charge is Q' and its voltage is V' = Q'/C, so the tiny work to add charge δQ' is δW = V' δQ' = (Q'/C) δQ'. Integrating from 0 to Q gives W = Q²/2C. The ½ is the average of a voltage that ramps linearly from 0 to V, i.e. average = V/2. So total work = Q × (V/2) = ½QV.
They are the exact same energy written three ways — pick whichever variables the question gives you. Use Q = CV to convert. Start from U = ½QV. Replace Q with CV → U = ½CV². Replace V with Q/C → U = Q²/2C. Rule of thumb: if the battery stays connected, V is fixed, so use ½CV². If the capacitor is charged then isolated (battery removed), Q is fixed, so use Q²/2C.
The energy is stored in the electric field in the gap between the plates, not 'on' the metal. This is why the same energy can be written as energy density ½ε₀E² multiplied by the volume Ad of the field region. For a parallel plate capacitor U = ½CV² = ½(ε₀A/d)(Ed)² = ½ε₀E²(Ad). The field carrying the energy is the key idea that leads to the next topic, energy density.
Yes. With the battery removed, Q is fixed. Energy is U = Q²/2C and C = ε₀A/d. Pulling plates apart increases d, so C decreases, so U increases. The extra energy comes from the work YOU do pulling the plates against their attraction. If instead the battery stays connected, V is fixed and U = ½CV² decreases as d increases (charge flows back to the battery).
A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and the area of each plate is A, the energy stored in the capacitor is (ε₀ = permittivity of free space):
Three identical capacitors P, Q and S, each of capacitance C, are connected to a battery of voltage V. P and Q are in series with each other, and S is directly across the battery. If the energy stored in capacitor P is U_P and the total energy stored in the system is U_T, then U_P/U_T is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
U = ½CV² = ½QV = Q²/2C, measured in joules. All three forms are equal; use Q = CV to switch between them depending on whether the question gives you C and V, Q and V, or Q and C.
Because voltage across the capacitor rises from 0 to V while charging, so charge is not pushed through the full V. The average voltage is V/2, making the stored work ½QV. QV would be correct only if the full voltage acted the whole time, which it does not.
If the battery stays connected, voltage V is constant → use ½CV². If the capacitor is charged and then the battery is disconnected, charge Q is constant → use Q²/2C. Choosing the right form saves algebra and avoids sign/direction errors when C changes.
No, but they are linked. Energy density is energy per unit volume of field, u = ½ε₀E². Total stored energy U = u × (volume Ad) for a parallel plate capacitor, which equals ½CV². Energy density is the next concept and explains WHERE the energy sits — in the field.
