Electrostatic Potential and Capacitance

Physics · Class 12

Lesson 6 of 13 · 7 min

Potential energy in an external field

NCERT §2.8

An electron breaks free from an earthed metal ball near Meera's dome and races towards the dome's surface at 300 kV. How much energy does it pick up? And what happens to the dipole stick placed between the charged plates?

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In short

Here the field comes from outside sources, which the charge in question is assumed not to disturb. Only the charge's energy in that field is counted, not the energy of the sources.

A single charge q at a point where the external potential is V(r) has potential energy qV(r).

Electron volt: the energy gained by an electron moved through 1 V. 1 eV = 1.6 × 10⁻¹⁹ J; 1 keV = 1.6 × 10⁻¹⁶ J, 1 MeV = 1.6 × 10⁻¹³ J.

Two charges in an external field: U = q₁V(r₁) + q₂V(r₂) + q₁q₂/(4πε₀r₁₂), the energy of each in the outside field plus their mutual energy.

Worked value: 7 μC and −2 μC held 18 cm apart on the x-axis, 9 cm either side of the origin, have mutual energy −0.7 J, so separating them completely takes 0.7 J. Put them in an outside field E = A/r² (A = 9 × 10⁵ N C⁻¹ m², V = A/r): the charges gain 70 J and −20 J, and the total is 70 − 20 − 0.7 = 49.3 J.

Dipole in a uniform field E: U(θ) = −pE cos θ = −p·E, taking U = 0 at θ = 90°. Turning it from θ₀ to θ₁ takes work pE(cos θ₀ − cos θ₁).

U is least (−pE) when p is along E, the stable position, and greatest (+pE) when p is opposite to E, an unstable one.

Worked value: a mole of dipoles, each 10⁻²⁹ C m, fully aligned in 10⁶ V m⁻¹ has U = −6 J. Turn the field by 60° and U rises to −3 J as the dipoles realign, so 3 J comes out as heat.

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