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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:__________

User Somenxavier
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3.3k points

2 Answers

21 votes
21 votes

Answer:

The electric field is E/8.

Step-by-step explanation:

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


E =(\rho r)/(3)

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

For case I:


\rho = (Q)/((4)/(3)\pi R^3)

So, electric field at a distance r is


E = \frac { 3 Q r}{3* 4\pi R^3}\\\\E = (Q r)/(4\pi R^3)

Case II:


\rho = (Q)/((4)/(3)\pi 8R^3)

So, the electric field at a distance r is


E' = \frac { 3 Q r}{3* 32\pi R^3}\\\\E' = (Q r)/(8* 4\pi R^3)\\\\E' = (E)/(8)

User Jon Preece
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2.9k points
21 votes
21 votes

Answer:

Hence the answer is E inside
= KQr_(1) /R^(3).

Step-by-step 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.

User Mkurz
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2.8k points