1a. If a charge Q is spread uniformly throughout the volume of an insulating sph

Question

1a. If a charge Q is spread uniformly throughout the volume of an insulating sphere (that is, charge cannot move around inside it) of radius r, what fraction of the charge lies within a radius of r/2? Note: The answer is not 1/2

1b. Explain why the dot product E*dA can be replaced with the expression EdA inside the integral where E is the magnitude of the electric field and dA is the magnitude of an element of area on the surface of the spherical shell. Hint: What is the angle between the vector E and the vector dA at any point on the spherical surface?

Explanation / Answer

1a it will be equals to the volume ratio

volume of sphere with (R/2)/ Volume of sphere of radius R

(4/3)*pi*(R/2)^3/[(4/3)*pi*(R)^3]

=1/8

1b)

Angle between the vector E and vector dA at any point is 00

So E*dA= EdAcos0=EdA

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