Problem 11-33 Electric potential - Part 5 - A

The figure shows two parallel, oppositely charged, plates with a potential difference of $6.0\; V$ between them. The plates are separated by $1.1\; cm.$ If we define the potential of the negative plate to be zero,

(a) what is the potential of the positive plate?
(b) what is the potential half way between the plates?
(c) what is the potential at point $A,$ which is $0.28\; cm$ above the positive plate?

Accumulated Solution

$V \text{ (lower plate)} = 6\; V \\ V \text{(half way)} = 3\; V \\ V(0.28\; cm) = 6 \times (1.1 - 0.28)/1.1 = 4.5 \;V \\ E = 6V/1.1 \times 10^{-2} \; m = 5.5 \times 10^{-2} V/mv$

Correct.

So $E = 6V/1.1 \times 10^{-2} m = 5.5 \times 10^{-2} V/m$

The direction of the field is:

(A)   from the positive to the negative plate.

(B)   from the negative to the positive plate