To develop this problem, we will apply the concepts related to the Indian electric field from Coulomb's law. According to the values given we have to the electric field is
[tex]E = 70000N/C[/tex]
Distance from the center of the ball to the point of consideration is
[tex]r = \frac{10}{2}+3.5[/tex]
[tex]r = 8.5cm[/tex]
Now the electric field strength is
[tex]E = \frac{kq}{r^2}[/tex]
Here,
k = Coulomb's Constant
q = Charge
r = Distance
Rearrange to find the charge,
[tex]q = \frac{Er^2}{k}[/tex]
Replacing,
[tex]q = \frac{(70000)(8.5*10^{-2})^2}{9*10^9}[/tex]
[tex]q = 5.6194*10^{-8}[/tex]
[tex]q = 56.194nC[/tex]
Therefore the charge on the ball is 56.194nC
