A sphere of radius 2 a is made of a nonconducting material that has a uniform vo

Question

A sphere of radius 2a is made of a nonconducting material that has a uniform volume charge density . Assume that the material does not affect the electric field. A spherical cavity of radius a is now removed from the sphere as shown in the figure below. Show that the electric field within the cavity is uniform and is given by

Ex = 0

and

Ey = a/30.

(Suggestion: The field within the cavity is the superposition of the field due to the original uncut sphere, plus the field due to a sphere the size of the cavity with a uniform negative charge density .) Do this on paper.

Explanation / Answer

We know that for an area A enclosing a volume V: 0*EdA = *dV

where 0= Permittivity of the medium

=the charge density in the sphere.

For a uniform sphere of radius r the integral on the left is 0*E(r)*4**r²,and the integral on the right is*(4/3)**r³. The field is directed radially since the dot product EdA is zero for any tangential component.

0*E(r)*4*r² = *(4/3)*r³

E(r) = *(1/3)*r/0

We are concerned with the field along the axis joining the center of the sphere and the center of the cavity; set this as the x-axis. Since the fields are radial, then the fields along this axis are co-linear and add algebraically

The field from the cavity with internal charge density of - is E(rc) = -*(1/3)*rc/0, where rc is the distance from the center of the cavity. Note that in passing through the center of the cavity, the direction of the "cavity" field reverses. a is the radius of the cavity.

From r = 2*a to center of the cavity rc = r - a and the cavity's field is -*(1/3)*(r - a)/0; from r = a to the sphere's center rc = a - r and the cavity's field is +*(1/3)*(a - r)/0

The sum of the fields is

from r = a to r = 2*a E(r) = *(1/3)*r/0 - *(1/3)*(r - a)/0 = *(1/3)*a/0

from r = a to 0, E(r) = *(1/3)*r/0 + *(1/3)*(a - r)/0 = *(1/3)*a/0

Since these are the same, the overall field in the cavity on axis is then E(r) = *(1/3)*a/0

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