Question

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

Answer 1

Answer:

The magnitude of the electric field between the plates is half its initial value.

Explanation:

We know the electric field E = V/d where V = voltage applied and d = separation between plates.

Since V is constant and V = Ed,

So, E₁d₁ = E₂d₂ where E₁ = initial electric field at separation d₁, d₁ = initial separation of plates, E₂ = final electric field at separation d₂ and d₂ = final separation of plates.

So, E₂ = E₁d₁/d₂

Now, the distance between the plates is twice their original separation. Thus, d₂ = 2d₁

So, E₂ = E₁d₁/2d₁ = E₁/2

So, E₂ = E₁/2

Thus, the magnitude of the electric field between the plates is half its initial value.