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?

a) 360 V
b) 450 V
c) 22.5 V
d) 0 V

Answer 1

The potential is given by 449.7 V.

Step-by-step 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 = (q)/(\pi R^2)\\\\\sigma = (10* 10^(-9))/(3.14* 0.2* 0.2 )\\\\\sigma = 7.96* 10^(-8) C/m^2`

The electric potential is given by

`V=(\sigma)/(2\varepsilon 0)\left ( √(R^2 + x^2) - x \right )\\\\V = (7.96* 10^(-8))/(2* 8.85* 10^(-12))\left ( √(0.2^2 + 0.15^2) - 0.15 \right )\\\\V = 449.7 V`

Answer 2

b) 450 V

Step-by-step explanation:

We are given that

Total charge, q=10nC=
`10* 10^(-9) C`

`1nC=10^(-9)C`

Radius, r=20 cm=
`(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=(q)/(A)=(q)/(\pi r^2)`

`\sigma=(10*10^(-9))/(3.14* (0.2)^2)`

Where
`\pi=3.14`

`\sigma=7.96* 10^(-8)C/m^2`

Electric potential,
`V=(\sigma)/(2\epsilon_0)(√(x^2+r^2)-x)`

Where
`\epsilon_0=8.85* 10^(-12)`

Using the formula

`V=(7.96* 10^(-8))/(2* 8.85* 10^(-12))(√((0.15)^2+(0.2)^2)-0.15)`

`V=449.7 V\approx 450V`

Hence, option b is correct.