Question

A solid non-conducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a radius r (r < R) from the center of the sphere the electric field has a value E. If the same charge Q were distributed uniformly throughout a sphere of radius 2R the magnitude of the electric field at a radius r would be equal to:__________

Answer 1

Answer:  

Hence the answer is E inside $= KQr_{1} /R^{3}$.

Explanation:  

E inside $= KQr_{1} /R^{3}$  

so if r1 will be the same then  

E  $\begin{bmatrix}Blank Equation\end{bmatrix}$ proportional to 1/R3  

so if R become 2R  

E becomes 1/8 of the initial electric field.

Answer 2

Answer:

The electric field is E/8.

Explanation:

The electric field due to a solid sphere of uniform charge density inside it is given by

$E =\frac{\rho r}{3}$

where, $\rho$ is the volume charge density and r is the distance from the center.

For case I:

$\rho = \frac{Q}{\frac{4}{3}\pi R^3}$

So, electric field at a distance r is

$E = \frac { 3 Q r}{3\times 4\pi R^3}\\\\E = \frac{Q r}{4\pi R^3}$

Case II:

$\rho = \frac{Q}{\frac{4}{3}\pi 8R^3}$

So, the electric field at a distance r is

$E' = \frac { 3 Q r}{3\times 32\pi R^3}\\\\E' = \frac{Q r}{8\times 4\pi R^3}\\\\E' = \frac{E}{8}$