Question

a non-conducting sphere of radius 1m carrying net positive charge 18nc is enclosed by a concentric non-conducting thin spherical shell of radius 2m carrying net negative charge 20nc.determine the electric fields at the distance of 0.5m,1.5m and 2.5m ,respectively,from the centre of the sphere

Answer 1

Solution: The electric field inside a uniformly charged sphere is zero. Outside the sphere, it coincides with the field of point charge located in its center. These two theorems, which are valid for electrostatic problems, are the basis of our solution. The electric fields at the distance of $r_1=$ 0.5m which is inside all sphers is null $E_1=0$ . The electric fields at the distance of 1.5m equales the electric field of first shere:

(1) $E_2=k\frac{q_1}{r_2^2}$ , $k=\frac{1}{4\pi \epsilon_0}=9⋅10^9 N·m^2/C^2$ - Coulomb constant for vakuum [1], $r_2=1.5m$, $q_1=18\cdot 10^{-9}C$ is net charge of first sphere. Sustitute these to Coulomb's law for electric field (1) we have

(2) $E_2=9\cdot 10^9 \frac{18\cdot 10^{-9}}{1.5^2}N\cdot m^2\cdot C^{-2}\cdot C\cdot m^{-2}=72N/C$

The electric fields at the distance of 2.5m determined by the sum of the electric fields of the two spheres

(3) $E_3=k\frac {q_1}{r_3^2}+k\frac {q_2}{r_3^2}=k\frac {q_1+q_2}{r_3^2}$ Since the charge of second sphere is negative we have

$E_3=9\cdot 10^9 \frac {(18-20)\cdot 10^{-9}}{2.5^2}N/C=-9\cdot\frac{2}{2.5^2}N/C=-2.9N/C$

A negative sign here means that the electric field vector outside of both spheres is directed along the radius to their center.

Answer: The electric fields at the distance of $r_1=$ 0.5m, $r_2=$ 1.5m and $r_3=$ 2.5m , from the centre of the sphere are $E_1=0, E_2=72N/C, E_3=-2.9N/C$

[1 ]https://en.wikipedia.org/wiki/Coulomb%27s_law