Question

Refer to Fig. 16–32d. If the two charged plates were moved until they are half the distance shown without changing the charge on the plates, the electric field near the center of the plates would

(a) remain almost exactly the same.

(b) increase by a factor of 2.

(c) increase, but not by a factor of 2.

(d) decrease by a factor of 2.

(e) decrease, but not by a factor of 2.

Short answer

The correct answer is option (a), remain almost exactly the same.

Step-by-step solution

Understanding of electric field between the plates of the capacitor

The electric field in between the capacitor plates depends on the surface charge density and permittivity of free space.

The expression for the electric field is given as follows:

\(E = \frac{\sigma }{{{\varepsilon _0}}}\) … (i)

Here,\(\sigma \)is the surface charge density, and\({\varepsilon _0}\)is the permittivity of free space.

Determination of the electric field with the change in the separation of plates.

From equation (i), it is clear that the electric field is independent of the distance between the plates and dependent only on the charge density; there will not be any change in the electric field on changing the distance.

The electric field lines from each plate are parallel with uniform density, unlike point charges. Taking the plates closer together will not affect the electric field between them. The electric field near the center of the plates would remain the same.

Therefore, the correct answer is option (a).