Question

What is the electric potential 15 cm above the center of a uniform charge density disk of total charge 10 nC and radius 20 cm?

Answer 1

Answer:

The potential is given by 449.7 V.

Explanation:

radius of disc, R = 20 cm = 0.2 m

distance, x = 15 cm = 0.15 m

charge, q = 10 nC

surface charge density

$\sigma = \frac{q}{\pi R^2}\\\\\sigma = \frac{10\times 10^{-9}}{3.14\times 0.2\times 0.2 }\\\\\sigma = 7.96\times 10^{-8} C/m^2$

The electric potential is given by

$V=\frac{\sigma}{2\varepsilon 0}\left ( \sqrt{R^2 + x^2} - x \right )\\\\V = \frac{7.96\times 10^{-8}}{2\times 8.85\times 10^{-12}}\left ( \sqrt{0.2^2 + 0.15^2} - 0.15 \right )\\\\V = 449.7 V$

Answer 2

Answer:

b) 450 V

Explanation:

We are given that

Total charge, q=10nC=$10\times 10^{-9} C$

$1nC=10^{-9}C$

Radius, r=20 cm=$\frac{20}{100}=0.2m$

1 m=100 cm

x=15 cm=0.15 cm

We have to find the electrical potential 15 cm above  the center of a uniform charge density disk .

We know that

$\sigma=\frac{q}{A}=\frac{q}{\pi r^2}$

$\sigma=\frac{10\times10^{-9}}{3.14\times (0.2)^2}$

Where $\pi=3.14$

$\sigma=7.96\times 10^{-8}C/m^2$

Electric potential,$V=\frac{\sigma}{2\epsilon_0}(\sqrt{x^2+r^2}-x)$

Where $\epsilon_0=8.85\times 10^{-12}$

Using the formula

$V=\frac{7.96\times 10^{-8}}{2\times 8.85\times 10^{-12}}(\sqrt{(0.15)^2+(0.2)^2}-0.15)$

$V=449.7 V\approx 450V$

Hence, option b is correct.