Question

The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50×104 V/m . (a) What is the potential difference between the plates? (b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?

Answer 1

(a) 3000 V

For two parallel conducting plates, the potential difference between the plates is given by:

$\Delta V=Ed$

where

E is the magnitude of the electric field

d is the separation between the plates

Here we have:

$E=7.50\cdot 10^4 V/m$ is the electric field

d = 4.00 cm = 0.04 m is the distance between the plates

Substituting,

$\Delta V=(7.50\cdot 10^4 V/m)(0.04 m)=3000 V$

(b) 750 V

The potential difference between the two plates A and B is

$\Delta V = V_B - V_A = 3000 V$

Let's take plate A as the plate at 0 volts:

$V_A = 0 V$

The potential increases linearly going from plate A (0 V) to plate B (3000 V).

So, if the potential difference between A and B, separated by 4 cm, is 3000 V, then the potential difference between A and a point located at 1 cm from A is given by the proportion:

$3000 V : 4 cm = V(1 cm) : 1 cm$

and solving for V(1 cm) we find:

$V(1 cm)=\frac{(3000 V)(1 cm)}{4 cm}=750 V$