-->

Electric Charges And Fields

Question

Wired Faculty App

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2µF. The separation is reduced to half and it is filled with a dielectric substance of value of 2.8. The final capacity of capacitor is

15.6μF

11.2μF

19μF

22.4μF

Solution

11.2μF

$C = \frac{ε_{0} A}{d} \propto \frac{1}{d} Hence, \frac{C_{1}}{C_{2}} = \frac{d_{2}}{d_{1}} or \frac{2}{C_{2}} = \frac{0.2}{0.4} or C_{2} = \frac{0.4 \times 2}{0.2} = 4 μF Final capacitor is C = 4 \times k = 4 \times 2.8 = 11.2 μF$

For Daily Free Study Material Join wiredfaculty

Electric Charges And Fields

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2µF. The separation is reduced to half and it is filled with a dielectric substance of value of 2.8. The final capacity of capacitor is

15.6μF

11.2μF

19μF

22.4μF

Some More Questions From Electric Charges and Fields Chapter

X-rays are

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Electric Charges And Fields

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2µF. The separation is reduced to half and it is filled with a dielectric substance of value of 2.8. The final capacity of capacitor is 15.6μF 11.2μF 19μF 22.4μF

Some More Questions From Electric Charges and Fields Chapter

X-rays are

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of 0.4 cm is 2µF. The separation is reduced to half and it is filled with a dielectric substance of value of 2.8. The final capacity of capacitor is

15.6μF

11.2μF

19μF

22.4μF