A parallel plate capacitor with plate area A and plate separation d has a uniform electric field E between its plates. The electrostatic energy stored in it is U . A second parallel plate capacitor has a plate area 2A and a plate separation d 2 . If the uniform electric field between the plates of the second capacitor is 2E , what is the electrostatic energy stored in the second capacitor?

Step-by-step solution

The electrostatic energy stored in a parallel plate capacitor is given by the product of energy density and volume: U = 1 2 ₀ E^2 (A d) For the second capacitor, the new area is A' = 2A , the new separation is d' = d 2 , and the new electric field is E' = 2E . The energy stored in the second capacitor is: U' = 1 2 ₀ (E')^2 (A' d') Substituting the given values: U' = 1 2 ₀ (2E)^2 (2A) ( d 2 ) U' = 1 2 ₀ (4E^2) (A d) U' = 4 ( 1 2 ₀ E^2 A d ) U' = 4U Answer: 4U