A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor i

Question

A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is $C_{1}$ . When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_{1}$ and the rest of the area gets charge $Q_{2}$ . The electric field in the dielectric is $E_{1}$ and that in the other portion is $E_{2}$ . Choose the correct option/options, ignoring edge effect.

diagram

TestHub · Accepted Answer

Both portions share the same plate voltage and d: $E_{1} = E_{2} = V / d$ - (A) correct.

$C_{1} = \frac{K ε _{0} A}{3 d}, C_{2} = \frac{2 ε _{0} A}{3 d} \Rightarrow \frac{C}{C _{1}} = \frac{K + 2}{K}$ - (D) correct.

$\frac{Q _{1}}{Q _{2}} = \frac{C _{1}}{C _{2}} = \frac{K}{2}$ - (C) wrong.

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Options:(select one or more)

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