Answer:
The electric field potential (E) is 1.75 X 10⁵ N/C
Explanation:
Electric field potential (E) is force per unit charge.
Electric field potential (E) = force/charge = f/q, unit is Newton (N) per Coulomb (C); N/C
[tex]E = \frac{q}{4 \pi \epsilon r^2}[/tex]
where;
q is charge in coulomb (C)
ε is permittivity of free space = 8.8542 X 10⁻¹² C²/Nm²
[tex]k = \frac{1}{4 \pi \epsilon} = \frac{1}{4\pi (8.8542 X 10^{-12})} = 8.99 X10^9 Nm^2/C^2[/tex]
if a = 7 X 10⁻⁶ C/m⁴ and b = 1m
a in "C" = [tex]7 X 10^-{6} \frac{C}{m^4} X (1m)^4 = 7 X 10^-{6} C[/tex]
Electric field potential at r = 0.6m = [tex]\frac{ka}{r^2}[/tex]
E = [tex]\frac{(8.99 X10^9)(7X 10^{-6})}{0.6^2}[/tex] = 1.75 X 10⁵ N/C
Therefore, the electric field potential (E) is 1.75 X 10⁵ N/C
