At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and respectively. (Take the potential to be zero at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge

(a) 0.3074 m

(b) 0.17 nC

(c) The electric field is directed away from the charge.

Explanation:

Parameters given:

Electric Potential, V = 4.98V

Electric field magnitude, E = 16.2V/m

(a) The electric potential is given as:

V = kq/r

Where

k = Coulombs constant

q = charge

r = distance

Making q the subject of formula:

q = (V*r)/k 1

Electric field magnitude, E is given as:

E = kq/r²

Making q subject of the formula:

q = (E*r²)/k 2

Equating 1 and 2,

(V*r)/k = (E*r²)/k

=> r = V/E

r = 4.98/16.2

r = 0.3074 m

(b) Charge, q, from 1:

q = (V*r)/k

q = (4.98 * 0.3074)/(9 * 10^9)

q = 0.17 nC

(c) The electric field and charge have positive values, hence the electric field is directed away from the charge.

laticia draws the diagram below to show earth’s magnetic field.

answer

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