Electrostatic Potential and Capacitance
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Direct answer
Potential — energy per unit charge — is a scalar, so superposition is mere arithmetic with signs: V = kq/r from each point charge, added. The field then points down the steepest potential slope (E = -dV/dx), and equipotential surfaces stand perpendicular to field lines. A capacitor stores this energy as field, C = Q/V with ½CV^2 held between plates whose capacitance Kε0A/d rises when a dielectric fills the gap — and what that dielectric does next depends entirely on whether the battery stayed connected, which is the most examined idea in the chapter.
What you must remember
- Point-charge potential V = kq/r: positive charges raise the potential, negative lower it, and the zero sits at infinity; potentials add as scalars, unlike fields.
- Equipotential surfaces are perpendicular to field lines, never intersect, and moving a charge along one requires zero work; crowded equipotentials mean a strong field.
- Dipole potential V = kp cosθ/r^2; potential energy of two point charges U = k q1 q2/r (negative for attraction).
- Parallel-plate capacitor: C = ε0A/d; a dielectric of constant K multiplies C by K and reduces the internal field to E0/K.
- Combinations: series 1/C = 1/C1 + 1/C2 (result smaller than either); parallel C = C1 + C2; charges equal on series capacitors, voltages equal on parallel ones.
- Stored energy U = ½CV^2 = Q^2/(2C) = ½QV; energy density in an electric field u = ½ε0E^2.
- Capacitors sharing charge reach the common potential V = (C1V1 + C2V2)/(C1 + C2), and energy is always lost in the sharing unless the initial voltages matched.
- A dielectric polarises — induced surface charges partially cancel the applied field — which is the microscopic reason C rises by K.
Battery connected or disconnected?
Slide a dielectric slab into a charged parallel-plate capacitor and track the same four quantities under two different contracts. Contract one: the battery stays connected, so V is pinned. Capacitance becomes KC; charge Q = KCV climbs to K times its old value; stored energy U = ½CV^2 rises K-fold, with the battery supplying the difference. Contract two: the battery was disconnected first, so Q is pinned instead. Capacitance still becomes KC, but now V = Q/C drops to 1/K of before, and the stored energy U = Q^2/(2C) falls to 1/K — the slab is literally pulled into the gap, and the field does that work on it. Every "what happens to charge/voltage/field/energy when K is inserted" MCQ in NEET-UG is one of these two ledgers wearing different clothes; the first reading task is to spot which quantity the question has frozen. The same fork governs moving the plates apart or inserting a metal slab, and it rewards students who argue from C = Q/V rather than memorising case tables.
Where students slip
Because potential is a scalar, the potential at the midpoint between equal and opposite charges is zero while the field there is emphatically not — the two field contributions point the same way and add; "V = 0 implies E = 0" is the chapter's signature false statement, and its converse is equally false. Second, work done moves charge between two potentials: W = q(V2 - V1), with the from-infinity convention only a special case where V(∞) = 0. Third, in the series combination students average the capacitances; series capacitance is below the smallest member, governed by the larger gap it effectively creates. Fourth, when two charged capacitors are joined, the lost energy ΔU = C1C2(V1 - V2)^2/(2(C1 + C2)) vanishes as heat and radiation in the connecting wires even if the wires are ideal resistance — the "lost" energy question is a fixture. Finally, remember that inserting a conductor of thickness t in a capacitor effectively reduces the plate separation by t, raising C — a fact the exam has used to disguise the dielectric problem as a metal-slab problem.
Frequently asked questions
Why is electric potential a scalar while field is a vector?
Potential is work per unit charge, and work adds algebraically with sign; no direction is assigned to it, so superposition of point-charge potentials is plain arithmetic.
What does a dielectric do to a capacitor's capacitance?
It multiplies C by the dielectric constant K by reducing the internal field to E0/K through polarisation; charge-voltage behaviour then follows from which of Q or V is held fixed.
How do capacitances combine in series and in parallel?
Series: 1/C = 1/C1 + 1/C2, giving a value smaller than the smallest; parallel: C = C1 + C2, always additive.
Why is energy lost when two capacitors at different potentials are connected?
Charge redistributes until a common potential is reached, and the difference, C1C2(V1 - V2)^2/(2(C1 + C2)), dissipates in the connecting wires as heat and electromagnetic radiation.
What is the energy density of an electric field?
u = ½ε0E^2 — energy stored per unit volume in any field region, dielectric or vacuum, the field itself being the seat of the energy.