# Answer to Question #7986 in Electromagnetism for wes edwards

Question #7986

The volume of a sphere of radius a is charged with a constant volume charge density of p. Find the field of this spherical charge @ a distance d>>a.

Expert's answer

Let S be the sphere of radius d with center at center of the ball,

and E be

the electric dield at some point of S.

Due to symmetricity of the ball and

sphere the absolute value of E is the same at all points of S.

Moreover the

vector E is parallel to normal vector to S.

Hence the flux Phi through S is

equal to

Phi = integral_over_S (E dA) = E * Area_of_S = E * 4 pi *

d^2.

On the other hand, due to Gauss' law

Phi is equal to the full

charge Q inside S divided by eps_0:

Phi = Q/eps_0.

The ball is

uniformly charged, so

Q = p * Volume_of_a_ball = p * 4/3 * pi *

a^3.

Hence

Phi = E * 4 pi * d^2 = p * 4/3 * pi * a^3 /

eps_0

Therefore

E = (p * 4/3 * pi * a^3) / (4 pi * d^2 * eps_0

)

= a^3 * p / (3 * eps_0 * d^2)

Answer: E = a^3 * p / (3 *

eps_0 * d^2)

and E be

the electric dield at some point of S.

Due to symmetricity of the ball and

sphere the absolute value of E is the same at all points of S.

Moreover the

vector E is parallel to normal vector to S.

Hence the flux Phi through S is

equal to

Phi = integral_over_S (E dA) = E * Area_of_S = E * 4 pi *

d^2.

On the other hand, due to Gauss' law

Phi is equal to the full

charge Q inside S divided by eps_0:

Phi = Q/eps_0.

The ball is

uniformly charged, so

Q = p * Volume_of_a_ball = p * 4/3 * pi *

a^3.

Hence

Phi = E * 4 pi * d^2 = p * 4/3 * pi * a^3 /

eps_0

Therefore

E = (p * 4/3 * pi * a^3) / (4 pi * d^2 * eps_0

)

= a^3 * p / (3 * eps_0 * d^2)

Answer: E = a^3 * p / (3 *

eps_0 * d^2)

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