Question

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

Answer 1

Answer:

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:

       $E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)$

• 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:

        $Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C$

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

      $E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C$

• 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.