Answer:
[tex]E=2.88\times 10^5\ N/C[/tex]
Explanation:
It is given that,
The radius of the solid sphere, R = 50 cm = 0.5 m
Charge on the sphere, [tex]q=40\ \mu C=40\times 10^{-6}\ C[/tex]
We need to find the magnitude of the electric field r = 10.0 cm away from the center of the sphere. The electric field at point r away form the center of the sphere is given by :
[tex]E=\dfrac{kqr}{R^3}[/tex]
[tex]E=\dfrac{9\times 10^9\times 40\times 10^{-6}\times 0.1}{0.5^3}[/tex]
[tex]E=2.88\times 10^5\ N/C[/tex]
So, the electric field 10.0 cm away from the center of the sphere is [tex]2.88\times 10^5\ N/C[/tex]. Hence, this is the required solution.
