Question

A solid conducting sphere of radius 2.00 cm has a charge of 8.40 µC. A conducting spherical shell of inner radius 4.00 cm and outer radius 5.00 cm is concentric with the solid sphere and has a charge of −2.55 µC. Find the electric field at the following radii from the center of this charge configuration.

Answer 1

Answer: the electric field is equal to: E=0 for r<2 cm; E= -2.29*10^4/r^2 for 2cm<r<4cm; E=0 for 4cm<r<5 cm; E=5.26*10^4/r^2 for r>5 cm

the electric field in N/C units

Explanation: In order to find the electric field for all r values, we have to use the definition of electric field and Gaussian law.

In this sense, for r<2 cm as it is inside a conductor teh electric field is zero.

for 2cm< r< 4 cm we applied the field from a spherical charge distribution so by the Gaussian law we find the total charge inside the gaussian surface so

E.4π r^2= Q inside/ε0 = -2.55μC/ε0

Idem for other regions.

for 4 cm<r< 5 cm in the outsider conductor the E=0

Finally, for r>5 cm

E.4π r^2=Q inside/ε0=(8.40-2.55)μC/ε0