The electric field between square the plates of a parallel-plate capacitor has magnitude E.The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation,which is small compared to the dimensions of the plates.The magnitude of the electric field between the plates is now equal to
A) 4E.
B) 2E.
C) E.
D) E/2.
E) E/4.

Question

Quizplus · Accepted Answer

The electric field between the plates of a parallel-plate capacitor is given by E = V/d, where V is the potential difference across the plates and d is the separation distance. The voltage across the plates is constant, so when the plates are pulled apart to twice their original separation, the separation distance becomes 2d. Therefore, the new electric field is given by E' = V/(2d).

Dividing E' by E, we get:

E'/E = (V/(2d)) / (V/d) = 1/2

So, the magnitude of the electric field between the plates is now half of its original value. Therefore, the answer is D) E/2.

The Electric Field Between Square the Plates of a Parallel-Plate