Question

An air-filled capacitor with plate area A and plate separation d has capacitance \( C_0 \). A slab of dielectric constant K, area A and thickness \( \frac{d}{5} \) is inserted between the plates. The capacitance of the capacitor will become:

• A. \( \left[ \frac{4K}{5K+1} \right] C_0 \)
• B. \( \left[ \frac{K+5}{4} \right] C_0 \)
• C. \( \left[ \frac{5K}{4K+1} \right] C_0 \)
• D. \( \left[ \frac{K+4}{5K} \right] C_0 \)

Hint:

Think of the air-filled portion and the dielectric-filled portion as two separate capacitors placed one after another between the same two plates. Decide whether two capacitors arranged this way combine in series or in parallel, get that right first, then find the capacitance of each part using its own thickness before combining them.

Answer 1

The Correct Option is C

Step 1: The capacitance of a parallel plate capacitor without any dielectric is given by: \[ C_0 = \epsilon_0 \frac{A}{d} \] where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates. Step 2: When a dielectric material with dielectric constant \( K \) is inserted between the plates, the capacitance increases by a factor of \( K \), but only for the portion of the plate that is covered by the dielectric. In this case, the dielectric slab occupies \( \frac{d}{5} \) of the total plate separation \( d \). Step 3: The new capacitance is given by the sum of the capacitance of the dielectric-filled region and the air-filled region: \[ C = \frac{\epsilon_0 A}{d - \frac{d}{5}} + \frac{\epsilon_0 K A}{\frac{d}{5}} \] Simplifying this expression: \[ C = \frac{\epsilon_0 A}{\frac{4d}{5}} + \frac{\epsilon_0 K A}{\frac{d}{5}} = \frac{5\epsilon_0 A}{4d} + \frac{5K \epsilon_0 A}{d} \] Step 4: Now, factoring out the common term \( \frac{\epsilon_0 A}{d} \), we get: \[ C = \frac{\epsilon_0 A}{d} \left( \frac{5}{4} + 5K \right) = C_0 \left( \frac{5}{4} + 5K \right) \] Step 5: Therefore, the new capacitance is: \[ C = C_0 \left[ \frac{5K}{4K+1} \right] \] Thus, the capacitance of the capacitor becomes \( \left[ \frac{5K}{4K+1} \right] C_0 \).

Answer 2

Concept:

• When a dielectric slab is inserted between the plates so that it occupies only part of the gap, the arrangement acts like two capacitors connected in series, not added together.
• One capacitor is the dielectric-filled part with thickness equal to the thickness of the slab, and the other is the remaining air-filled part.
• For capacitors in series, the reciprocals of capacitance add up, not the capacitances themselves.

Step 1: Identify the two capacitors formed
The gap of width $d$ splits into two parts: a dielectric part of thickness $rac{d}{5}$ with dielectric constant $K$, and an air part of thickness $d - rac{d}{5} = rac{4d}{5}$. Both parts share the same plate area $A$.

Step 2: Write the capacitance of each part
Air part: $C_1 = rac{\epsilon_0 A}{4d/5} = rac{5\epsilon_0 A}{4d}$
Dielectric part: $C_2 = rac{K\epsilon_0 A}{d/5} = rac{5K\epsilon_0 A}{d}$

Step 3: Combine the two capacitors in series
$rac{1}{C} = rac{1}{C_1} + rac{1}{C_2} = rac{4d}{5\epsilon_0 A} + rac{d}{5K\epsilon_0 A}$
$rac{1}{C} = rac{d}{5\epsilon_0 A}\left(4 + rac{1}{K} ight) = rac{d}{5\epsilon_0 A} \cdot rac{4K+1}{K}$

Step 4: Invert to get the final capacitance
$C = rac{5K\epsilon_0 A}{d(4K+1)} = rac{5K}{4K+1} \cdot rac{\epsilon_0 A}{d}$
Since $C_0 = rac{\epsilon_0 A}{d}$, this becomes $C = \left[rac{5K}{4K+1} ight]C_0$

Final Answer: $C = \left[\dfrac{5K}{4K+1} ight] C_0$