Question

A capacitor, of capacitance $C$, plate area $A$ and plate separation $d$ is charged to $V$ volts by a battery. The energy stored in it is $U$. After some time, the battery is disconnected and a slab of dielectric constant $K$, is inserted to fill the space between the plates of the capacitor. Let $V'$ and $U'$ now be the potential difference across the plates of the capacitor and the energy stored in it respectively. Find $\left(\dfrac{V'}{V} ight)$ and $\left(\dfrac{U'}{U} ight)$.

Hint:

The battery is disconnected, so the charge stays constant while the capacitance becomes KC. Then V = Q/C and U = Q^2/2C both drop by the factor K.

Answer 1

Step 1: Understanding the Concept:
When the battery is disconnected, the charge on the plates cannot leave. So the charge \(Q\) stays constant. A dielectric of constant \(K\) changes the capacitance but not the charge.

Step 2: Key Formula or Approach:
1. Capacitance with the dielectric: \(C' = KC\).
2. Potential difference: \(V = Q/C\).
3. Energy stored: \(U = \frac{Q^2}{2C}\).

Step 3: Before inserting the slab:
The capacitor is charged to \(V\), so \(Q = CV\) and \(U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}\).

Step 4: Potential difference after the slab:
The charge is still \(Q\) and the capacitance is \(KC\).
\[ V' = \frac{Q}{KC} = \frac{V}{K} \quad \Rightarrow \quad \frac{V'}{V} = \frac{1}{K} \]

Step 5: Energy after the slab:
\[ U' = \frac{Q^2}{2KC} = \frac{U}{K} \quad \Rightarrow \quad \frac{U'}{U} = \frac{1}{K} \]
The energy falls because the slab is pulled into the gap by the field. The field does positive work on the slab, so the stored energy is lower.

Final Answer:
Both the potential difference and the stored energy are reduced by the factor \(K\). \[ \boxed{\frac{V'}{V} = \frac{1}{K}, \qquad \frac{U'}{U} = \frac{1}{K}} \]