Get Answers to all your Questions

header-bg qa

A capacitor has capacitance 5 \mu \mathrm{F} when it's parallel plates are separated by air medium of thickness \mathrm{d}. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness \frac{d}{2} is inserted between the plates. Capacitance of the capacitor in the presence of slab will be \mu \mathrm{F}.

Option: 1

6


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{C}_{\mathrm{o}}=\frac{\in_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}=5

\begin{aligned} & \mathrm{C}_1=\frac{\epsilon_{\mathrm{o}}(1.5) \mathrm{A}}{\frac{\mathrm{d}}{2}}=3 \mathrm{C}_0, \mathrm{C}_2=\frac{\epsilon_{\mathrm{o}} \mathrm{A}}{\frac{\mathrm{d}}{2}}=2 \mathrm{Co} \\ & C_{eq}=\frac{3 \mathrm{C}_{\mathrm{o}} \times 2 \mathrm{C}_{\mathrm{o}}}{5 \mathrm{C}_{\mathrm{o}}}=\frac{6}{5} \times 5=6 \mu \mathrm{f} \end{aligned}

Posted by

Anam Khan

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE