Capacitance

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Question

Plates of a parallel plate capacitor are connected to the terminals of a battery. If the separation between the plates is doubled, then the electric field at a point in the space between the plates

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Solution

initial capacitance $C = \frac{ε_{0} A}{d}$
Initial charge, $Q = CV, E = \frac{σ}{ε_{0}} = \frac{Q}{ε_{0} A} = \frac{C V}{ε_{0} A}$
$Finally, C' = \frac{ε_{0} A}{2 d} = \frac{C}{2}, Q' = C' V = \frac{C V}{2}, E' = \frac{σ'}{ε_{0}} = \frac{Q'}{A ε_{0}} = \frac{C V}{2 ε_{0} A} = \frac{E}{2}$

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Capacitance

Remember concepts with our Masterclasses.

Plates of a parallel plate capacitor are connected to the terminals of a battery. If the separation between the plates is doubled, then the electric field at a point in the space between the plates

initial capacitance C=ε0AdInitial charge, Q=CV, E=σε0=Qε0A=CVε0AFinally, C'=ε0A2d=C2, Q'=C'V=CV2, E'=σ'ε0=Q'Aε0=CV2ε0A=E2

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