The potential difference between the plates of a parallel plate capacitor is 4.00 V. If the plate separation for this capacitor

Question

Ivanshal [37]

3 years ago

12

The potential difference between the plates of a parallel plate capacitor is 4.00 V. If the plate separation for this capacitor

is 6.00 cm, what is the intensity of the electric field between the plates of this capacitor?

Physics

1 answer:

shtirl [24] · Answer 1 · 2022-02-12T04:28:13+03:00

<h2>Answer:</h2>

66.67 V / m

<h2>Explanation:</h2>

The electric field intensity (E) between the plates of a capacitor is the ratio of the potential difference (V) between the plates to the distance (d) of separation between the plates. Electric field intensity can be measured in V/m. It can be written as follows;

E = V / d ----------------------------(i)

<em>From the question;</em>

V = 4.00V

d = 6.00cm = 0.06m

<em>Substitute these values into the equation above. i.e equation (i) as follows;</em>

E = 4.00 / 0.06

E = 66.67 V / m

Therefore, the electric field intensity between the plates of the capacitor is 66.67 V / m