Question

A parallel plate capacitor having area \(A\) and separated by distance \(d\) is filled by a copper plate of thickness \(b\). The new capacity is:

• A. \(\dfrac{\varepsilon_0 A}{d+\frac{b}{2}}\)
• B. \(\dfrac{\varepsilon_0 A}{2d}\)
• C. \(\dfrac{\varepsilon_0 A}{d-b}\)
• D. \(\dfrac{2\varepsilon_0 A}{d+\frac{b}{2}}\)

Hint:

Remember:

• Conducting slab inside capacitor: \[ d_{\text{effective}}=d-b \]
• New capacitance: \[ C=\frac{\varepsilon_0 A}{d-b} \]
• Inserting conductor increases capacitance because effective separation decreases.

Answer 1

The Correct Option is C

Concept: When a conducting slab is inserted between the plates of a capacitor, the electric field inside the conductor becomes zero. Hence, the effective separation between capacitor plates reduces.

Step 1: Determine effective plate separation. Original separation: \[ d \] Thickness of copper slab: \[ b \] Since electric field inside conductor is zero, effective air gap becomes: \[ d-b \]

Step 2: Use capacitance formula. Capacitance of a parallel plate capacitor is given by: \[ C=\frac{\varepsilon_0 A}{d} \] Replacing effective separation: \[ d \rightarrow d-b \] Therefore: \[ C=\frac{\varepsilon_0 A}{d-b} \] Therefore, the correct answer is: \[ \boxed{\frac{\varepsilon_0 A}{d-b}} \]