Question

A parallel plate capacitor is connected with a battery. If the plates of capacitor are shifted apart, then which of the following is the correct statements.

• A. The intensity of electric field between the plates decreases and charge on plates also decreases
• B. The intensity of electric field between the plates remains constant but charge on the plates increases
• C. The intensity of electric field between the plates remains constant but charge on the plates decreases
• D. The intensity of electric field between the plates increases but charge on the plates decreases

Answer 1

The correct option is A The intensity of electric field between the plates decreases and charge on plates also decreases

The capacitance of a parallel plate capacitor is given by the formula $$C=\frac{{\epsilon }_{0}A}{d}$$ Where,
$A$ is the area of each plate,
$d$ is the distance between the plates and
${\epsilon }_0$ is the permittivity of the space between the plates.

So, with increase in separation between the plates, the capacitance will decrease.

In the presence of the battery, potential difference $V$ remains constant and from $Q=CV$, we can observe that the charge will also decrease.

As we know that, electric field intensity beween the parallel plates is given by $$E=\frac{Q}{A{\epsilon }_0}$$As, $Q$ is decreasing here, so $E$ will also decrease.

Hence, option (a) is the correct answer.

Key Concept: When the battery remains connected, the charge on the plates can change.