Question

The charge density of radius 12.0 cm isgiven by with a = 1.40 μC/m 3 and r ismeasured radially outward from theorigin. What is the electric potential atpoint A, a distance 40.0 cm above thedisk? Hint: you will need to integrate thenon uniform charge distribution to findthe electric potential.

Answer 1

Answer:

The electric potential of the uniformly charge disk is 1392.1 V

Explanation:

Electric potential, for a uniformly charged disk at a distance A, is given as;

$V = \frac{\sigma}{2 \epsilon} [\sqrt{A^2 +R^2} -A]$

Where;

σ is the charge density = 1.40 μC/m³

ε is the permittivity of free space = 8.85 x 10⁻¹²

A is the distance above the disk = 40 cm = 0.4 m

R is the radius of the disk = 0.12 m

Substitute in these values into the equation above, we will have

$V = \frac{1.4 X 10^{-6}}{2X8,85X10^{-12}}[\sqrt{0.4^2 +0.12^2}-0.4] \\\\V = (79096.05)(0.0176) = 1392.1 V$

Therefore, the electric potential of the uniformly charge disk is 1392.1 V