A parallel plate capacitor has plates of area $$A$$ and separation $$d$$ and is charged to a potential difference$$V$$. The charging battery is then disconnected and the plates are pulled apart until their separation is 2$$d$$. What is the work required to separate the plates?

Question

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

$$2{{\varepsilon }_{0}}A{{V}^{2}}/d$$

TuteeHUB · Answer

$${{\varepsilon }_{0}}A{{V}^{2}}/d$$

TuteeHUB · Answer

$$3{{\varepsilon }_{0}}A{{V}^{2}}/d$$

TuteeHUB · Answer

$${{\varepsilon }_{0}}A{{V}^{2}}/2d$$

1.	A parallel plate capacitor has plates of area \[A\] and separation \[d\] and is charged to a potential difference\[V\]. The charging battery is then disconnected and the plates are pulled apart until their separation is 2\[d\]. What is the work required to separate the plates?
A.	\[2{{\varepsilon }_{0}}A{{V}^{2}}/d\]
B.	\[{{\varepsilon }_{0}}A{{V}^{2}}/d\]
C.	\[3{{\varepsilon }_{0}}A{{V}^{2}}/d\]
D.	\[{{\varepsilon }_{0}}A{{V}^{2}}/2d\]
Answer» E.

A parallel plate capacitor has plates of area \[A\] and separation \[d\] and is charged to a potential difference\[V\]. The charging battery is then disconnected and the plates are pulled apart until their separation is 2\[d\]. What is the work required to separate the plates?

Discussion

No Comment Found

Related MCQs

Reply to Comment