An isolated parallel-plate capacitor has plate area and initial separation

PHYSICS · Multi-part · CBSE · Physics · Electrostatic Potential and Capacitance · Electrostatic Potential and Capacitance

Problem

An isolated parallel-plate capacitor has plate area

A

and initial separation

d_{1}

. The plates carry equal and opposite charges of magnitude

Q_{0}

. The plates are then slowly pulled apart to a new separation

d_{2}

(

d_{2} > d_{1}

), with no charge allowed to leave the plates at any time. Throughout the process, edge effects remain negligible. (a) Obtain an expression for the capacitance as a function of plate separation

d

for this capacitor. (b) Derive an expression for the electric-potential energy stored in the capacitor as a function of

d

, keeping

Q_{0}

constant. (c) Using your result from part (b), derive a symbolic expression for the mechanical work that must be done by an external agent to change the separation from

d_{1}

d_{2}

. Express your answer in terms of

Q_{0}

A

\epsilon_{0}

d_{1}

, and

d_{2}

. (d) Evaluate the work required if

A = 0.020\,\text{m}^2

Q_{0} = 8.0\times10^{-6}\,\text{C}

d_{1} = 1.0\,\text{mm}

, and

d_{2} = 4.0\,\text{mm}

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