A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andouter radius 5 cm is concentric with the solid sphere and has acharge of -4 microCoulomb. What is the magnitude and direction ofthe electric field at r = 7 cm?

A)0

B)1 N/C radially outward

C)7.35 106 N/C radiallyoutward

D)7.35 106 N/C radiallyinward

E)3.56 107 N/C radiallyoutward

F)8 107 N/C radially outward

C) 7.35*10⁶ N/C radially outward

Explanation:

If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.So, we can write the following equation:

       

As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.So, on the outer surface of the shell there must be a charge that be the difference between them:

       

Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

     

As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.

The speed was not exceeded so it was slow