Question

A dielectric is inserted between the plates of a capacitor connected to a battery. What happens?

• A. Charge decreases
• B. Electric field increases
• C. Electric field decreases
• D. Charge increases

Hint:

If battery remains connected: \[ V = \text{constant} \] If battery is disconnected: \[ Q = \text{constant} \] Always check whether the battery is connected or disconnected in capacitor problems.

Answer 1

The Correct Option is D

Concept: When a dielectric material is inserted between the plates of a capacitor connected to a battery:

• Capacitance increases
• Potential difference remains constant
• Charge stored increases

The relation is: \[ Q = CV \] where:

• \(Q\) = charge
• \(C\) = capacitance
• \(V\) = potential difference

Step 1: Understand the effect of dielectric on capacitance.
Capacitance of a parallel plate capacitor is: \[ C = \frac{\varepsilon_0 A}{d} \] After inserting dielectric of dielectric constant \(K\): \[ C' = K C \] Thus capacitance increases.

Step 2: Analyze the condition of battery connection.
Since capacitor remains connected to the battery: \[ V = \text{constant} \] Battery continuously maintains the same potential difference.

Step 3: Apply the charge relation.
Using: \[ Q = CV \] Since:

• \(C\) increases
• \(V\) remains constant

Therefore: \[ Q \uparrow \] Hence charge stored on the capacitor increases.

Step 4: Choose the correct option.
Thus, the correct answer is: \[ \boxed{\text{Charge increases}} \]