Question

If a = 7 × 10−6 C/m4 and b = 1 m, find E at r = 0.6 m. The permittivity of a vacuum is 8.8542 × 10−12 C 2 /N · m2 . Answer in units of N/C

Answer 1

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

$E = \frac{q}{4 \pi \epsilon r^2}$

where;

q is charge in coulomb (C)

ε is permittivity of free space = 8.8542 X 10⁻¹² C²/Nm²

$k = \frac{1}{4 \pi \epsilon} = \frac{1}{4\pi (8.8542 X 10^{-12})} = 8.99 X10^9 Nm^2/C^2$

if a = 7 X 10⁻⁶ C/m⁴  and b = 1m

a in "C" = $7 X 10^-{6} \frac{C}{m^4} X (1m)^4 = 7 X 10^-{6} C$

Electric field potential at r = 0.6m = $\frac{ka}{r^2}$

E = $\frac{(8.99 X10^9)(7X 10^{-6})}{0.6^2}$ = 1.75 X 10⁵ N/C

Therefore, the electric field potential (E) is  1.75 X 10⁵ N/C