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

Answer:

Therefore,

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

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

$radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm$

$r = 0.145\ cm = 0.145\times 10^{-2}\ m$

Charge ,

$Q = 30.0\ pC=30\times 10^{-12}$

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

$V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}$

Where,  

V= Electric potential,  

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

Q = Charge  

r = Radius  

Substituting the values we get

$V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}$

$V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt$

Therefore,

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