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Question

Two parallel-plate capacitors, identical except that one has twice the plate separation of the other, are charged by the same voltage source. Which capacitor has a stronger electric field between the plates? Which capacitor has a greater charge? Which has greater energy density? Explain your reasoning.

Solution

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Answered 1 year ago
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Given

We are given two parallel-plate capacitors, identical except that one has twice the plate separation this means the next

C1=C2=CC_{1} = C_{2} = C

V1=V2=VV_{1} = V_{2} = V

And d1=dd_{1} = d while d2=2dd_{2} = 2d.

Required

Which capacitor has a stronger electric field EE, charge QQ and energy density uu

Explanation

The electric field depends on the separated distance between the two plates and it is given by

E=Vd\begin{equation} E = \dfrac{V}{d} \end{equation}

As shown by equation (1), the electric field is inversely proportional to the separated distance dd as the distance increases the electric field decreases.

If we solve equation (1) to E1E_{1} and E2E_{2} we could get the ratio between the two electric fields as next

E1E2=(V/d1)(V/d2)E1E2=(V/d)(V/2d)E1E2=2E1=2E2\begin{gather*} \dfrac{E_{1}}{E_{2}} = \dfrac{(V/d_{1})}{(V/d_{2})}\\ \dfrac{E_{1}}{E_{2}} = \dfrac{(V/d)}{(V/2d)} \\ \dfrac{E_{1}}{E_{2}} = 2 \\ E_{1} = 2E_{2} \end{gather*}

Therefore, the capacitor with smaller separated distance has a greater electric field.

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