Answer:
[tex]q=2.08*10^{-8}C\\ q=20.8nC[/tex]
Explanation:
Given data
Electric potential V=1.50 kV
Diameter d=25.0 cm
radius=diameter/2=25/2=12.5 cm=0.125 m
to find
We are asked to find excess charge
Solution
As inside the sphere the electric field is zero everywhere and potential is same at every point inside the sphere and on its surface
So we can use the value of the potential to get the charge q on the sphere
[tex]V=\frac{1}{4\pi E}\frac{q}{R}\\ q=4\pi ERV\\Where\\4\pi E=\frac{1}{9.0*10^{9}N.m^{2}/C^{2} }\\ q=(\frac{1}{9.0*10^{9}N.m^{2}/C^{2} })(0.125m)(1.50*10^{3}V )\\q=2.08*10^{-8}C\\ q=20.8nC[/tex]
