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frutty [35]
3 years ago
8

A solid nonconducting sphere of radius R carries a uniform charge density throughout its volume. At a radial distance r1 = R/4 f

rom the center, the electric field has a magnitude E0.
What is the magnitude of the electric field at a radial distance r2 = 4R?

A) E0/2

B) 4E0

C) E0

D) E0/4

E) 2E0
Physics
1 answer:
aleksandr82 [10.1K]3 years ago
8 0

Answer:D

Explanation:

If Q is the total charge on the sphere then

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

Charge enclosed up to r=\frac{R}{4}

q_{enclosed}=\frac{Q}{\frac{4}{3}\pi R^3}\times \frac{4}{3}\pi (\frac{R}{4})^3

q_{enclosed}=\frac{Q}{64}

Applying Gauss law

EA=\frac{q_{enclosed}}{\epsilon }

E=\frac{\frac{Q}{64}}{4\pi (\frac{R}{4})^2}

E=\frac{1}{4}\times \frac{Q}{4\pi \epsilon }

and Electric field at r=\frac{R}{4}

E_0=\frac{1}{4}\times \frac{Q}{4\pi \epsilon }

at r=4 R

charge enclosed is Q

applying Gauss law

E\cdot A=\frac{q_{enclosed}}{\epsilon }

E=\frac{q_{enclosed}}{4\pi (4R)^2\epsilon }

E=\frac{1}{4}\times \frac{Q}{4\pi \epsilon }\times \frac{1}{4}

E=\frac{E_0}{4}

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