Answer:
b) 450 V
Explanation:
We are given that
Total charge, q=10nC=[tex]10\times 10^{-9} C[/tex]
[tex]1nC=10^{-9}C[/tex]
Radius, r=20 cm=[tex]\frac{20}{100}=0.2m[/tex]
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
[tex]\sigma=\frac{q}{A}=\frac{q}{\pi r^2}[/tex]
[tex]\sigma=\frac{10\times10^{-9}}{3.14\times (0.2)^2}[/tex]
Where [tex]\pi=3.14[/tex]
[tex]\sigma=7.96\times 10^{-8}C/m^2[/tex]
Electric potential,[tex]V=\frac{\sigma}{2\epsilon_0}(\sqrt{x^2+r^2}-x)[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]
Using the formula
[tex]V=\frac{7.96\times 10^{-8}}{2\times 8.85\times 10^{-12}}(\sqrt{(0.15)^2+(0.2)^2}-0.15)[/tex]
[tex]V=449.7 V\approx 450V[/tex]
Hence, option b is correct.
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
[tex]\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[/tex]
The electric potential is given by
[tex]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[/tex]
