Answer:
Therefore,
The net charge enclosed within this surface is
[tex]Q=2.31\times 10^{-9}\ C[/tex]
Explanation:
Given:
Radius, r = 0.4 m
Electric field at the sphere's surface
E = 130 N/C
To Find:
Charge, Q = ?
Solution:
Electric field at the sphere's surface is given as,
[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r^{2}}[/tex]
Where,
E = Electric Field ,
ε0 = permeability free space = 8.85 × 10–12 F/m
Q = Charge
r = Radius
Substituting the values we get
[tex]130=\dfrac{1}{4\pi\times 8.85\times 10^{-12}}\times \dfrac{Q}{(0.4)^{2}}[/tex]
[tex]Q=20.8\times 4\times 3.14\times 8.85\times 10^{-12}=2.31\times 10^{-9}\ C[/tex]
Therefore,
The net charge enclosed within this surface is
[tex]Q=2.31\times 10^{-9}\ C[/tex]
