Physics · Electrostatic Potential And Capacitance · NEET
When two charged conductors are connected by a wire, charge flows from the higher-potential body to the lower-potential body until BOTH reach the SAME (common) potential V. Charge is redistributed so that it splits in the ratio of their capacitances (Q proportional to C), NOT their radii of size. Memory hook: "Water levels equalise" — like water in two connected tanks flows until the levels match, charge flows until the potentials match.
Two spheres joined by a wire share charge until both sit at the same potential V. Charge splits as Q proportional to R (bigger sphere takes more charge), but surface charge density sigma = Q/(4 pi R^2) is higher on the smaller sphere.
Your doubts, answered
When two conductors touch, does the charge split equally between them?
No. Charge splits in the ratio of their CAPACITANCES, so Q proportional to C. Only if the two conductors are identical (same capacitance) does the charge split equally. For two spheres, C proportional to radius R, so the bigger sphere takes MORE charge (Q1/Q2 = R1/R2). Equal-halves thinking is the number-one mistake.
What decides the direction charge flows — higher charge or higher potential?
Higher POTENTIAL, not higher charge. Charge always flows from the body at higher potential to the body at lower potential until both potentials become equal. A small body with a lot of charge can be at higher potential than a big body with the same charge, so charge can flow OUT of the smaller one. Never compare charges to decide direction; compare V = Q/C.
What is the common potential and how do I find it?
When joined, total charge is conserved and shared over total capacitance. Common potential V = (Q1 + Q2)/(C1 + C2) = (C1V1 + C2V2)/(C1 + C2). This is just total charge divided by total capacitance. After sharing, each body sits at this same V.
Is energy conserved when two conductors are connected?
No. Charge is conserved, but electrostatic energy is always LOST (unless the potentials were already equal). The lost energy is dissipated as heat in the connecting wire (and as tiny radiation/spark). Energy loss = (1/2)(C1 C2)/(C1 + C2) times (V1 - V2) squared. This is a favourite NEET trap: students assume energy stays the same.
Two spheres, one big one small, are connected — which ends up at higher surface charge density?
The SMALLER sphere. Both reach the same potential, so Q proportional to R, but surface charge density sigma = Q/(4 pi R squared) proportional to 1/R. So the smaller radius has the larger sigma. This is exactly why sharp points (very small R) leak charge easily — the basis of lightning rods and corona discharge.
⚠️ The NEET trap ✗ When a charged capacitor is connected to an identical uncharged one, the total stored energy stays the same because charge is conserved. ✓ Charge IS conserved, but energy is NOT. Total capacitance doubles to 2C while charge Q is fixed, so common V halves. U_final = (1/2)(2C)(V/2)^2 = (1/4)CV^2 = U_initial/2. Half the energy is lost as heat. 🧠 Charge conserved does NOT mean energy conserved. Whenever two conductors at different potentials are joined, energy is always lost as heat in the wire.
Real NEET questions
NEET 2017
A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system:
A · Increases by a factor of 4
B · Decreases by a factor of 2 ✓
C · Remains the same
D · Increases by a factor of 2
Solution: Battery removed, so charge Q is fixed (conserved). Connecting an identical uncharged capacitor in parallel doubles the total capacitance to 2C. Since Q is unchanged, common potential drops: V' = Q/(2C) = V/2. Initial energy U_i = (1/2)CV^2. Final energy U_f = (1/2)(2C)(V/2)^2 = (1/2)(2C)(V^2/4) = (1/4)CV^2 = U_i/2. So the energy DECREASES by a factor of 2 (the other half is lost as heat in the connecting wire). Answer: B.
NEET 2021
Two charged spherical conductors of radius R1 and R2 are connected by a wire. Then the ratio of surface charge densities of the spheres (sigma1/sigma2) is:
A · R1/R2
B · R2^2/R1^2
C · R1^2/R2^2
D · R2/R1 ✓
Solution: When connected by a wire, both spheres reach a COMMON potential. For a sphere V = kQ/R, so kQ1/R1 = kQ2/R2, giving Q1/Q2 = R1/R2 (bigger sphere takes more charge). Surface charge density sigma = Q/(4 pi R^2). Therefore sigma1/sigma2 = (Q1/Q2) x (R2^2/R1^2) = (R1/R2) x (R2^2/R1^2) = R2/R1. The smaller sphere has the higher charge density. Answer: D.
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 common potential formula for two connected conductors?
V = (Q1 + Q2)/(C1 + C2) = (C1V1 + C2V2)/(C1 + C2). Total charge is conserved and is shared over the total capacitance.
Why is energy lost when two conductors are connected?
Charge moving through the connecting wire faces resistance, so some electrostatic energy is converted to heat (and a little radiation). Energy is lost whenever the two bodies start at different potentials. Loss = (1/2)(C1C2/(C1+C2))(V1 - V2)^2.
Do two connected spheres end up with equal charge?
Only if they have equal radius. In general they reach equal POTENTIAL, so charge divides as Q proportional to R (Q1/Q2 = R1/R2). The larger sphere carries more charge.
Which sphere has higher surface charge density after connection?
The smaller sphere. Since sigma = Q/(4 pi R^2) and Q proportional to R, we get sigma proportional to 1/R. Smaller radius means larger sigma — this is why sharp points discharge easily.
Why does this concept matter for NEET?
NEET repeatedly tests common potential, energy loss (percentage dissipated), and the surface-charge-density ratio of connected spheres. Knowing that charge follows capacitance (not equal split) and that energy is always lost saves you from the two classic traps.
