Question

Consider a parallel-plate capacitor with plates of area A and with separation d.
Part A
Find F(V), the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor.
Express your answer in terms of given quantities and ϵ0.
View Available Hint(s)for Part A


F(V)

Answer 1

The magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as $\frac{V^2 A \varepsilon _o }{2d^2}$.

Magnitude of the force

The magnitude of the force each plate experiences due to the other plate is determined as follows;

F = U/d

where;

• U is potential energy stored in the capacitor

$F = \frac{1}{2} \frac{Q^2}{C} \times \frac{1}{d}$

Q = CV

$F = \frac{1}{2} \frac{C^2V^2}{C d} = \frac{CV^2}{2d}$

where;

• C is the capacitance

The capacitance is given as;

$C = \frac{\varepsilon _o A }{d}$

$F = \frac{\varepsilon _o A }{d} \times \frac{V^2}{2d} \\\\F = \frac{V^2 A \varepsilon _o }{2d^2}$

Thus, the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as $\frac{V^2 A \varepsilon _o }{2d^2}$.

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