# Electric field vectors for two circular parallel plates

See attached file for clarity.

Consider two circular parallel plates of radius R separated by a distance d  R. Assume that these plates have uniformly distributed surface charges of Q and -Q. Electric field vectors in the region between the plates will all point perpendicular to the plates from the positive plate to the negative plate and will have a magnitude of 4k

a) Use equation d=dVe/d=-qE-dr/q= d=-Edr

And the information just given to evaluate the potential difference between the plates in terms of Q, R, and d.

b) Argue that you get the same results using the formula for the potential of an infinite plate :

-s= 2k E=2kr

## Solution This solution is **FREE** courtesy of BrainMass!

a.)

Because,

Electric field strength is given as:

E = 4pi*k*s

here, s == sigma = Q/(pi*R^2)

k= 1/(4pi*e) (e stands for epsilon).

therefore,

E = 4*pi*k*(Q/(pi*R^2))

dV = -E.dr =

-ve sign shows that potential will decrease along to the direction of E.

=> dV = 4*pi*k*(Q/(pi*R^2))*d (where d is the separation between plates)

=> dV = 4k*d*Q/R^2 --Answer

Positive plate will be at higher potnetila and -ve at lower potential and their potential diff will be = 4k*d*Q/R^2 --Answer

b.)

Because,

potential due to a single infinite plate = 2pi*k*s*r

therefore,

because of two plates = 2*2pi*k*s*d = 4pi*k*s*d = 4k*d*Q/R^2

--Answer