Question

The potential difference between the accelerating plates of a TV set is about 25 kV. If the distance between the plates is 1.0 cm, find the magnitude of the uniform electric field in the region between the plates.

Answer 1

Answer:

E = 2.5 x 10⁶ N/C

Explanation:

V = Potential difference between the plates of a TV set = 25 kV = 25000 Volts

d = Distance between the plates of TV set = 1.0 cm = 0.01 m

E = Electric field in the region between the plates

Electric field between the plates is given as

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

Inserting the values

$E= \frac{25000}{0.01}$

E = 2.5 x 10⁶ N/C

Answer 2

The uniform electric field is the field in which the value of the electric field strength remain same at each point.

The magnitude of the uniform electric field in the region between the plates  $2.5\times10^6 \rm N/C$.

What is uniform electric field?

The uniform electric field is the field in which the value of the electric field strength remain same at each point.

It can be given as,

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

Here, $V$ is the  potential difference between two points and $d$ is the distance between two points.

Given information-

The  potential difference between the accelerating plates of a TV set is about 25 k-V or 25000 V.

The distance between the plates is 1.0 cm or 0.01 meters.

Use the above formula to find the magnitude of the uniform electric field in the region between the plates as,

$E=\dfrac{25000}{0.01}\\E=2.5\times10^6 \rm N/C$

Thus the magnitude of the uniform electric field in the region between the plates  $2.5\times10^6 \rm N/C$.

Learn more about uniform electric field here;

https://brainly.com/question/14632877