Answer
1.2 X $10^{-9}$ C = q
r - 0.07 m
Work Step by Step
The electric field and potential are related as
V = Er.
Given :
V = 155 V
E = 2240 N/C. So
155 = 2240 r
r = 0.07 m.
Electric potential due to a point charge is given by
V = k$\dfrac{q}{r}$, where
r = distance from point charge= 0.07 m,
k = 9.0 X $10^{9} Nm^{2} / C^{2}$, and
q = charge. So
155 = 9.0 X $10^{9} \frac{q}{0.07}$
1.2 X $10^{-9}$ C = q.
