Answer:
Therefore,
The potential (in V) near its surface is 186.13 Volt.
Explanation:
Given:
Diameter of sphere,
d= 0.29 cm
[tex]radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm[/tex]
[tex]r = 0.145\ cm = 0.145\times 10^{-2}\ m[/tex]
Charge ,
[tex]Q = 30.0\ pC=30\times 10^{-12}[/tex]
To Find:
Electric potential , V = ?
Solution:
Electric Potential at point surface is given as,
[tex]V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}[/tex]
Where,
V= Electric potential,
ε0 = permeability free space = 8.85 × 10–12 F/m
Q = Charge
r = Radius
Substituting the values we get
[tex]V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}[/tex]
[tex]V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt[/tex]
Therefore,
The potential (in V) near its surface is 186.13 Volt.
