Question

. Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the electric field strength between them, if the potential 8.00 cm from the zero volt plate (and 2.00 cm from the other) is 450 V? (b) What is the voltage between the plates?

Answer 1

Answer:

Explanation:

electrical field strength = potential diff / distance

= ( 450 - 0 ) / 8 x 10⁻²

= 5625 V/m

b )

potential difference = electric field x distance

= 5625 x .10

= 562.5 V