Question

Determine the rate at which the electric field changes between the round plates of a capacitor, 5.8 cm in diameter, if the plates are spaced 1.2 mm apart and the voltage across them is changing at a rate of 150 V/s.

Answer 1

Answer:

The rate at which the electric field changes between the round plates of a capacitor is $125\times 10^{3}Vs^{-1}$.

Explanation:

It is given in the problem that the round plates of a capacitor are spaced some distance apart and the voltage across them is changing.

The expression for the electric field in terms of voltage is as follows;

$E=\frac{V}{d}$

Here, E is the electric field, V is the voltage and d is the distance of separation.

Differentiate expression of the electric field with respect to time, t.

$\frac{dE}{dt}=\frac{1}{d}\frac{dV}{dt}$

Convert the distance of separation from mm to m.

d= 1.2 mm

$d=1.2\times 10^{-3}m$

Calculate the rate at which the electric field changes.

$\frac{dE}{dt}=\frac{1}{d}\frac{dV}{dt}$

Put $\frac{dV}{dt}=150 Vs^{-1}$ and $d=1.2\times 10^{-3}m$

$\frac{dE}{dt}=\frac{1}{1.2\times 10^{-3}}(150)$

$\frac{dE}{dt}=125\times 10^{3}Vs^{-1}$

Therefore, the rate at which the electric field changes is $125\times 10^{3}Vs^{-1}$.