## Related questions with answers

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

Verified#### Given

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

$C_{1} = C_{2} = C$

$V_{1} = V_{2} = V$

And $d_{1} = d$ while $d_{2} = 2d$.

#### Required

Which capacitor has a stronger electric field $E$, charge $Q$ and energy density $u$

#### Explanation

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

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

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

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

$\begin{gather*} \dfrac{E_{1}}{E_{2}} = \dfrac{(V/d_{1})}{(V/d_{2})}\\ \dfrac{E_{1}}{E_{2}} = \dfrac{(V/d)}{(V/2d)} \\ \dfrac{E_{1}}{E_{2}} = 2 \\ E_{1} = 2E_{2} \end{gather*}$

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

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