A solid nonconducting sphere has a radius of 50.0 cm. It has a uniform charge de

Question

A solid nonconducting sphere has a radius of 50.0 cm. It has a uniform charge density of 3.00×10-3 C/m3. What is the strength of the electric field (a) at a distance of 2.00 m from the center of the sphere? (b) at the surface of the sphere? (c) at a distance of 25.0 cm from the center of the sphere? (Hint: use the shell theorem, which states that when calculating an electrostatic force or field outside a sphere with a uniform charge, the sphere can be treated as though all of its charge resides at the center.)

Explanation / Answer

since the sphere is non conducting the charge is uniformly distributed throughout the volume of the sphere.
charge densiy = 3 e-3 C / m ^-3
(a) since the point is outside the sphere so the net electric field is as if entire charge were concentrated at its centre.
total charge= volume density * volume =( 3 e -3 ) * ( 4/3) ( 0 .5 ^ 3) = 1.57 e -3 C

so the electric field is (1/4e0) * ( q/ r ^2) =( 9 e 9 ) * ( 1.57 e -3 ) / ( 2 ^ 2) = 3.52 e 6    SI units

( b ) just at the surface of the sphere the electric field is given by (1/4e0) * ( q / r ^2 )

= ( 9 e 9 ) * ( 1.57 e -3 / .5 ^2) = 5.65 e 7

(c) shell theorem means that the electric field at a point in shell with uniform charge density is equal to the total charge inside the sphere with radius equal to the distnace of that point from the centre.

so the total charge that will exert force = ( 3 e - 3 ) * ( 4/3) ( .25 ^ 3) = 1.96 e - 4 C

so the electric field is (1/4e0) * ( q/ r ^2) =( 9 e 9 ) * ( 1.96 e -4 ) / ( .25 ^ 2) = 2.822 e 7

hence the answers

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