Question

A solid conducting sphere with radius R = 0.390 m carries a net charge of +0.650 nC.

Electric field: Find the electric field: at a point 0.100 m outside the surface of the sphere.

Answer 1

38.5 N/C

Step-by-step explanation:

The electric field generated by a charged sphere at a point outside the sphere is equivalent to the electric field generated by a single point charge, and it is given by

`E=k(Q)/(r^2)`

where

`k=9\cdot 10^9 Nm^2C^(-2)` is the Coulomb's constant

Q is the net charge

r is the distance from the centre of the sphere

In this problem, we have

`Q=+0.650 nC=0.65\cdot 10^(-9)C`

`r=0.390 m`

Substituting into the equation, we find

`E=(9\cdot 10^9)((0.65\cdot 10^(-9)C))/((0.390m)^2)=38.5 N/C`