Answer:
E = 1.20 × [tex]10^{4}[/tex] N/C
V = 1800 V
x = 0.058 m
Explanation:
given data
radius = 15 cm
net charge = 3 × [tex]10^{-8}[/tex] C
electric potential decreased = 500 V
solution
we get here electric field at the sphere’s surface that is
electric field at the sphere’s surface E = [tex]\frac{q}{4\pi \epsilon _o R^2}[/tex] ............1
put here value
electric field at the sphere’s surface E = [tex]\frac{3\times 10^{-8}\times 8.99\times 10^9}{ 0.15^2}[/tex]
E = 1.20 × [tex]10^{4}[/tex] N/C
and
potential on surface of sphere is
V = [tex]\frac{q}{4\pi \epsilon _o R}[/tex] ................2
V = [tex]\frac{3\times 10^{-8}\times 8.99\times 10^9}{ 0.15}[/tex]
V = 1800 V
and
now we get distance that is x
and we know here
ΔV = V(x) - V ..............3
substitute here value
-500V = [tex]\frac{q}{4\pi \epsilon _o } \times (\frac{1}{R+x} - \frac{1}{R})[/tex]
-500 V = [tex]{3\times 10^{-8}\times 8.99\times 10^9} \times (\frac{1}{0.15+x} - \frac{1}{0.15})[/tex]
solve it we get x
x = 0.058 m
