A solid, insulating sphere of radius a = 2 cm has a total charge Q= 10 nC which

Question

A solid, insulating sphere of radius a = 2 cm has a total charge Q= 10 nC which is uniformly distributed. Using Gauss's Law determine the magnitude of the electric field at a point inside sphere which is 1 cm from its center. ( The test was supposed to be at r = 1 cm instead of 4 cm as given on the actual exam) at r= 1 cm, E = 1.13 times 103 N/C at r = 4 cm, E = 5.6 times 104 N/C An electric dipole has opposite charges of 5.0 times 10-15 C which are separated by a distance of 0.50 mm. It is oriented in a uniform electric field of 1.0 times 103 N/C. If the magnitude of the torque exerted on the dipole by the electric field is 1.05 times 10-15 Nm, what is its orientation with respect to the electric field? 25degree In the figure, point A is 30 cm below the ceiling. Determine how much longer it will take for a pulse to travel along wire 1 than it would along wire 2. Both wires are made of the same material and have identical cross sections.

Explanation / Answer

Let a = 5 cm, b = 10 cm and c = 15 cm. also let q = 3 uC. Now the charge density in the conductiong sphere is

p = 3q/(4*pi*a^3)

Consider 0 < r < a. Guass' law gives

Integral(E*dA) = q'/eo =1/e0 * integral(p*dV) = 4*pi*p/e0 * integral(r^2dr) = 4*pi*r^3*p/(3*e0) = q/e0*(r/a)^3

Integral(E*dA) = E*4*pi*r^2 = q/e0 * (r/a)^3 ---> E = q*r/(4*pi*e0*a^3)


Now consider a < r < b and apply Gauss' law:

Integral (E*dA) = q/e0 ---> E*4*pi*r^2 = q/e0 ----> E = q/(4*pi*e0*r^2)

When r = b, E = E(b) = q/(4*pi*e0*b^2)

For b< r < c, you are inside the conductor and the field is by defnition 0 ----> E = 0

For r > c Gauss's law gives

E*4*pi*r^2 = (q+q1)/e0 where q1 = -1 uC

E = (q+q1)/(4*pi*e0*r^2)

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