Problem 11-33 Electric potential - Part 5 - B

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?

Diagram of parallel charged plates.

Accumulated Solution

\(V \text{ (lower plate)} = 6\; V \\ V \text{(half way)} = 3\; V \\ V(0.28\; cm) = 6 \times (1.1 - 0.28)/1.1 = 4.5 \;V\)

No. Just knowing the units of the electric field will give you the answer here. The units of \(E\) are \(V/m\) (or \(J/C\)), therefore the dimensionally correct answer is:

(A)   \(E = \Delta V/d\)

(B)   \(E = \Delta Vd\)

(C)   \(E = d/\Delta V\)