CLASS 12 · CHAPTER 2 · ELECTROSTATICS
Electrostatic Potential & Capacitance
Charge up a capacitor, then disconnect the battery and pull the plates apart. The voltmeter reading shoots up. Did the field between the plates get stronger too? It didn't move at all.
Watch it happen
Battery connected: V is held fixed at 100 V. Drag d and watch Q, E, and stored energy all drop as the plates separate.
With the battery connected, dragging the plates apart drops the voltage reading, the charge readout, and the stored energy all together, exactly as intuition expects. Flip the switch to disconnect the battery first, then drag the same slider: now the charge is trapped, so the voltage climbs sharply as the plates separate, yet the field-line thickness between the plates, the actual field strength, doesn’t change at all.
Where the formula comes from
Electrostatic potential at a point is the work done per unit charge to bring a small positive test charge from infinity to that point. For a single point charge :
Capacitance of any conductor (or pair of conductors) is simply the ratio of charge stored to the potential difference that charge creates, a fixed property of the geometry alone:
For two parallel plates of area , separated by a small gap , applying Gauss’s law to find the uniform field between them (), then integrating across the gap, gives:
Charging a capacitor means moving charge onto one plate against an ever-growing opposing voltage, so the energy stored is the area under a Q-V graph, not simply :
Where the shortcut stops working
The entire chapter hinges on one question that’s easy to answer wrong on reflex: when the plates of a capacitor move apart, what stays fixed, the charge or the voltage? The answer depends entirely on whether the battery is still connected, and the two cases behave completely differently.
Battery connected: the battery enforces at a fixed value no matter what you do to . Since falls as grows, and with pinned, charge must flow back through the battery as shrinks. Everything, , , and , drops together.
Battery disconnected: now has nowhere to go, it’s trapped on the plates, fixed. Since depends only on the charge density, not the gap, the field cannot change as you separate the plates. But , and is now fixed while grows, so must climb in direct proportion. The energy climbs too, paid for entirely by the mechanical work of pulling two attracting plates apart.
Students who memorise “moving plates apart changes the capacitor” without pinning down which quantity is actually held fixed get this wrong close to half the time on an exam, precisely because the two cases send and in opposite directions.
Apply it under exam conditions
Q1. A parallel-plate capacitor with plate area 200 cm² and separation 2 mm is connected to a 50 V battery. Find its capacitance and the charge stored.
Q2. The same capacitor is then disconnected from the battery and its separation is doubled to 4 mm. Find the new voltage and the new field strength, and compare the field to its original value.
Disconnected means stays at 4.43 nC. The new capacitance is half the old one (since doubled):
But the field only ever depended on the fixed charge density:
Check it directly: disconnect the battery in the simulation above at 2 mm, then drag the slider to 4 mm, the voltage readout doubles while the field-line thickness doesn’t shift.
Quick answers
If I pull a charged, isolated capacitor's plates apart, does the field between them get weaker, since the plates are farther apart now?+
No, and this is the chapter's sharpest trap. With the charge trapped (battery disconnected), the field between the plates is E = σ/ε₀, fixed entirely by the charge density on the plates, which doesn't change as you move them apart. What does change is the voltage, V = Ed, since d itself is growing while E stays put. It's a rare case in CBSE physics where the 'obvious' answer (field weakens with distance) is simply wrong, because this isn't a point charge's field falling off with distance, it's a uniform field between two plates, which doesn't depend on the gap at all.
Where does the extra energy come from when I pull the plates apart on a disconnected capacitor?+
From you. The two oppositely charged plates attract each other, so pulling them apart means doing mechanical work against that attraction, and that work has to go somewhere: it becomes extra stored electrical energy in the capacitor, U = Q²/2C, which increases as C decreases. Energy is conserved, it's just converted from your muscular effort into field energy.
Is electrostatic potential the same thing as electric field?+
No, and mixing them up is one of the most common mistakes in this chapter. The field E is a vector, defined at every point, and tells you the force per unit charge. The potential V is a scalar, and tells you the potential energy per unit charge. They're related, E is literally the (negative) rate of change of V with position, but V existing at a point doesn't mean E is large there, and vice versa, only how fast V is changing matters for E.
Why does connecting two capacitors in series give a SMALLER capacitance than either one alone?+
Because connecting capacitances in series is mathematically identical to adding resistances in parallel, 1/C_series = 1/C1 + 1/C2, which always gives a result smaller than the smallest individual C. Physically: the same charge Q has to sit on every capacitor in the series chain, so the voltages add up (V = Q/C1 + Q/C2 + ...) for a fixed total Q, meaning the combination needs more total voltage to hold the same charge than any single capacitor would, i.e. the overall C = Q/V_total is smaller.
Related concepts
Physics doesn’t stay inside chapter boundaries. Neither should you.