What will be the change in the electric field of the parallel plate capacitor when a dielectric -of constant k- is pulled out from between the plates of the capacitor and it remains connected to the battery-

Question

Toppr Admin · Accepted Answer

Correct option is D- decreases by a factor of 1-kThe electric field is given by -E-dfrac-Q-K-epsilon-0A-Therefore- -E- reduces by a factor of -1-K-

What will be the change in the electric field of the parallel plate capacitor when a dielectric (of constant k) is pulled out from between the plates of the capacitor and it remains connected to the battery.

Correct option is D. decreases by a factor of 1/k
The electric field is given by $$E=\dfrac{Q}{K\epsilon_0A}$$
Therefore, $$E$$ reduces by a factor of $$1/K$$

What will be the change in the electric field of the parallel plate capacitor when a dielectric (of constant k) is pulled out from between the plates of the capacitor and it remains connected to the battery.

Correct option is D. decreases by a factor of 1/kThe electric field is given by $$E=\dfrac{Q}{K\epsilon_0A}$$Therefore, $$E$$ reduces by a factor of $$1/K$$

Correct option is D. decreases by a factor of 1/k
The electric field is given by $$E=\dfrac{Q}{K\epsilon_0A}$$
Therefore, $$E$$ reduces by a factor of $$1/K$$