If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance
Text Solution
Verified by ExpertsC
In order to obtain high capacitance, plats of large area should be taken and kept close to each other.
The capacitance of a parallel plate capacitor of plate area A and the distance d, between the plated is given by
$c = \frac{K \epsilon_o A}{d}$ Given, $d_1 = d$, $d_2 = \frac{d}{2}$, $K_1 = K$, $K_2 = 2K$ $\frac{C_1}{C_2} = \frac{K_1}{d_1} \times \frac{d_2}{K_2}$ $= \frac{K}{d} \times \frac{\frac{d}{2}}{2 \times K}$ $\frac{C_1}{C_2} = \frac{1}{4}$ $\Rightarrow C_2 = 4 C_1$ Hence, capacitance increases four times.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems