Answer:
(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.
