Question

A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on its surface. What is the potential (in V) near its surface?

Answer 1

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Step-by-step explanation:

Given:

Diameter of sphere,

d= 0.29 cm

`radius=(d)/(2)=(0.29)/(2)=0.145\ cm`

`r = 0.145\ cm = 0.145* 10^(-2)\ m`

Charge ,

`Q = 30.0\ pC=30* 10^(-12)`

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

`V=(1)/(4\pi\epsilon_(0))* (Q)/(r)`

Where,

V= Electric potential,

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge

r = Radius

Substituting the values we get

`V=(1)/(4* 3.14* 8.85* 10^(-12))* (30* 10^(-12))/(0.145* 10^(-2))`

`V=(30)/(16.117* 10^(-2))=186.13\ Volt`

Therefore,

The potential (in V) near its surface is 186.13 Volt.