Question

The electric field strength is 1.50×104 N/C inside a parallel-plate capacitor with a 1.10 mm spacing. An electron is released from rest at the negative plate?

Answer 1

The strength of the electric field between two parallel conducting plates separated by a distance of 1.10 mm is 1.50x10^4 N/C.

Step-by-step explanation:

The strength of an electric field between two parallel conducting plates separated by a distance of 1.10 mm can be calculated using the formula:

Electric Field Strength (E) = Voltage (V) / Distance (d)

Given that the electric field strength is 1.50 x 10^4 N/C and the distance is 1.10 mm, we can rearrange the formula to solve for the voltage:

Voltage (V) = Electric Field Strength (E) x Distance (d)

Substituting the given values, we get:

Voltage (V) = (1.50 x 10^4 N/C) x (1.10 x 10^-3 m)

Simplifying the calculation gives the voltage as 16.5 V.