Question

A parallel plate capacitor of capacitance \(5\ \mu F\) and plate separation \(6\ \text{cm}\) is connected to a \(1\ \text{V}\) battery and charged. A dielectric of dielectric constant \(4\) and thickness \(4\ \text{cm}\) is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is:

• A. \(2\ \mu C\)
• B. \(3\ \mu C\)
• C. \(5\ \mu C\)
• D. \(10\ \mu C\)

Hint:

Remember: \[ d_{\text{eq}}=(d-t)+\frac{t}{K} \]

• Dielectric insertion increases capacitance
• If battery remains connected: \[ Q=CV \] changes because \(C\) changes

Answer 1

The Correct Option is C

Concept: When a dielectric slab is partially inserted between capacitor plates, the effective separation becomes: \[ d_{\text{eq}}=(d-t)+\frac{t}{K} \] where: \[ d = \text{plate separation} \] \[ t = \text{thickness of dielectric slab} \] \[ K = \text{dielectric constant} \] New capacitance is: \[ C' = \frac{\varepsilon_0 A}{d_{\text{eq}}} \]

Step 1: Find equivalent separation. Given: \[ d=6\ \text{cm} \] \[ t=4\ \text{cm} \] \[ K=4 \] Therefore: \[ d_{\text{eq}} = (6-4)+\frac{4}{4} \] \[ d_{\text{eq}} = 2+1 = 3\ \text{cm} \]

Step 2: Find new capacitance. Original capacitance: \[ C=5\ \mu F \] Since capacitance is inversely proportional to separation: \[ \frac{C'}{C} = \frac{d}{d_{\text{eq}}} = \frac{6}{3} = 2 \] Thus: \[ C' = 2\times5 \] \[ C' = 10\ \mu F \]

Step 3: Calculate additional charge. Battery voltage remains constant: \[ V=1\ \text{V} \] Initial charge: \[ Q=CV = 5\times1 = 5\ \mu C \] Final charge: \[ Q'=C'V = 10\times1 = 10\ \mu C \] Additional charge: \[ \Delta Q = Q'-Q \] \[ \Delta Q = 10-5 = 5\ \mu C \] Therefore, the correct answer is: \[ \boxed{5\ \mu C} \]