Question

When a dielectric slab of dielectric constant \( k = 2 \) is used to fill the space between the plates of a parallel plate capacitor, the capacitance of the capacitor is found to be \( 20 \mu F \). What will be the capacitance when the slab is replaced with air?

• A. \( 10 \mu F \)
• B. \( 40 \mu F \)
• C. \( 20 \mu F \)
• D. \( 5 \mu F \)

Hint:

Capacitance increases by factor \( k \) when dielectric is inserted. Removing dielectric divides capacitance by \( k \).

Answer 1

The Correct Option is A

Step 1: Write relation of capacitance with dielectric.
\[ C = k C_0 \]

Step 2: Identify given values.
\[ k = 2,\quad C = 20\mu F \]

Step 3: Substitute values.
\[ 20 = 2C_0 \]

Step 4: Solve for \( C_0 \).
\[ C_0 = 10\mu F \]

Step 5: Interpret result.
\( C_0 \) is capacitance without dielectric (i.e., air).

Step 6: Final comparison.
Capacitance reduces when dielectric is removed.

Step 7: Final Answer.
\[ \boxed{10\mu F} \]