A parallel plate capacitor has plates of area $0.01\ \text{m}^2$ separated by $4\ \text{mm}$. A dielectric slab of dielectric constant $K=6$ and thickness $2\ \text{mm}$ is inserted so as to partly fill the gap. Find (a) the new capacitance, and (b) the charge stored if the capacitor is connected to a $100\ \text{V}$ battery.

A parallel plate capacitor has plates of area 0.01m² separated by 4mm. A dielectric slab of dielectric constan

Numerical · Q24

0.01\ \text{m}^2

(a) the new capacitance, and

(b) the charge stored if the capacitor is connected to a

100\ \text{V}

battery.

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Given $A=0.01\ \text{m}^2$ , $d=4\times 10^{-3}\ \text{m}$ , $t=2\times 10^{-3}\ \text{m}$ , $K=6$ :

(d-t) + \frac{t}{K} = (2\times 10^{-3}) + \frac{2\times 10^{-3}}{6} = 2\times 10^{-3} + 3.33\times 10^{-4} = 2.333\times 10^{-3}\ \text{m}

C = \frac{\epsilon_0 A}{(d-t)+t/K} = \frac{(8.85\times 10^{-12})(0.01)}{2.333\times 10^{-3}} \approx 3.79\times 10^{-11}\ \text{F} \approx 37.9\ \text{pF}

(b) With $V=100\ \text{V}$ : …