Electric potential at a point in an electric field is the work done in bringing a unit positive test charge from infinity to that point without acceleration. It is a scalar quantity. V = W / q0, and its SI unit is the Volt (V = J/C).
Key Concepts
- Electric Potential due to a point charge: V = kq / r. For positive q, V is positive; for negative q, V is negative.
- Potential due to a system of charges: V = k sum(qi / ri) — algebraic sum, since V is a scalar.
- Equipotential surfaces: Surfaces on which the potential is the same at every point. The electric field is always perpendicular to equipotential surfaces. No work is done in moving a charge along an equipotential surface.
- Relation between E and V: E = -dV/dr (along the field direction). The field points from high potential to low potential.
- Electric Potential Energy: The work done in assembling a configuration of charges. For two charges: U = kq1q2 / r. For a charge q in an external potential V: U = qV.
- Conductors in electrostatic equilibrium: The electric field inside is zero; charge resides on the surface; the entire conductor (including surface) is an equipotential; the field just outside is sigma/epsilon0 perpendicular to the surface.
- Capacitance: A capacitor stores electric charge and energy. C = Q/V. Its SI unit is the Farad (F). A parallel plate capacitor: C = epsilon0 A / d.
- Effect of dielectric: When a dielectric of dielectric constant K is inserted, capacitance becomes C = K epsilon0 A / d.
- Energy stored in a capacitor: U = Q2 / (2C) = CV2 / 2 = QV/2.
- Capacitors in series: 1/Ceq = 1/C1 + 1/C2 + 1/C3 ... (same charge on each).
- Capacitors in parallel: Ceq = C1 + C2 + C3 ... (same voltage across each).
- Key Formulas
- V = kq / r (point charge)
- C = Q / V; C = epsilon0 A / d (parallel plate)
- U = CV2 / 2 = Q2 / (2C)
- Cseries: 1/Ceq = sum(1/Ci)
- Cparallel: Ceq = sum(Ci)
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Find the potential at a distance 0.2 m from a +5 µC charge.
V = kq/r = (9 × 109 × 5 × 10-6) / 0.2 = 45000 / 0.2 = 2.25 × 105 V
Two charges +2 µC and -2 µC are 0.4 m apart. Find the potential at the midpoint.
r1 = r2 = 0.2 m
V = k(+2 × 10-6)/0.2 + k(-2 × 10-6)/0.2 = 0 V (potentials cancel as they are equal and opposite)
A parallel plate capacitor has plates of area 0.01 m2 separated by 0.002 m. Find the capacitance.
C = epsilon0 A / d = 8.85 × 10-12 × 0.01 / 0.002 = 4.425 × 10-11 F = 44.25 pF
Three capacitors of 2 µF, 3 µF, and 6 µF are connected in series. Find the equivalent capacitance.
1/C = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1
Ceq = 1 µF
Three capacitors of 2 µF, 4 µF, and 6 µF are in parallel. Find Ceq and charge on each if connected to 12 V.
Ceq = 2 + 4 + 6 = 12 µF
Q1 = 2 × 12 = 24 µC; Q2 = 4 × 12 = 48 µC; Q3 = 6 × 12 = 72 µC
A 10 µF capacitor is charged to 100 V. Find the energy stored.
U = CV2 / 2 = 10 × 10-6 × 1002 / 2 = 10 × 10-6 × 10000 / 2 = 0.05 J
The capacitance of a parallel plate capacitor is 5 µF with air. A dielectric of K = 3 is inserted. Find the new capacitance. If the charge remains constant, find the new voltage.
C' = KC = 3 × 5 = 15 µF. Original V = Q/C. New V' = Q/C' = Q/(3C) = V/3. Voltage reduces to one-third.
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Common mistakes
- Electric potential is a scalar — when adding potentials due to multiple charges, use algebraic (signed) addition, not vector addition.
- Energy stored in a capacitor (CV2/2) should not be confused with energy supplied by the battery (CV2 — battery supplies double the energy stored; half is dissipated as heat).
- Capacitance depends only on geometry and dielectric, NOT on charge or voltage.
Summary
Electric potential quantifies the electrical state at a point and is key to understanding energy in electrostatics. Capacitors store charge and energy, with series and parallel combinations giving different equivalent capacitances. Dielectrics enhance capacitance by reducing the effective field.