Answer to Question #91101 in Electricity and Magnetism for Promise Omiponle

Question

Two small spheres spread 10 cm apart have equal but opposite charge. If the magnitude of the force between them is 2.5 x 10^-10 N
(a)Calculate the charge on each sphere.
(b)Calculate the electric field intensity at the mid-point.
(c)Calculate the electric potential at the mid-point.
(d) How many excess electrons must be present on each sphere.

Answer 1

a) According to Coulomb's law for two equal charges:

$F=\frac{1}{4\pi\epsilon_0}\cdot\frac{q^2}{r^2},$

$q=2r\sqrt{\pi\epsilon_0F}=1.667\cdot10^{-11}\text{ C}.$

b) The field at the midpoint is a sum of fields of two charges. Since they have opposite signs, in the middle the field is twice as the field of a single charge (because the field is a vector quantity):

$E=2\cdot\frac{1}{4\pi\epsilon_0}\cdot\frac{q}{(r/2)^2}=119.86\text{ V/m}.$

c) The electric potential is 0 in the middle since the two charges are equal but opposite:

$V=\frac{1}{4\pi\epsilon_0}\cdot\frac{q}{r/2}-\frac{1}{4\pi\epsilon_0}\cdot\frac{q}{r/2}=0.$

d) It's easy to calculate by dividing the total charge of a sphere by the charge of an electron:

$N=\frac{q}{e}=104050000.$