Question

the electric field due to a point charge is 40N/C and the electric potential at that point is 20j/C. Calculate the distance of the point from the charge and also find the magnitude of the charge.

Answer 1

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E = kq/r2 = 40 N/C

and

electric Potential


V = kq/r = 20 J/C


now,

V/E = (kq/r) / (kq/r2) = 20 / 40

or

V/E = r = 1/2

thus,

the distance will be

r = 0.5m

and also......

V2/kE = q

or

q = 202 / (9x109 x 40)

or

q = (400 x 10-9) / 9x40

thus,

the magnitude of charge will be

q = 1.11 nC

Answer 2

E=kq/r^2 =40N/C
and V=kq/r=20N/C
V/E=(kq/r2)/(kq/r^2)=1/2
Or,r=0.05m