Charge Distribution on Connected Spheres

## Question:

There are two charges: 1 positively charged sphere (holding +
10nC) with redius 1 cm & 1 negatively charged sphere (holding
- 2 nC) with radius 2 cm. Their centres are seperated by 1 m. The
two spheres are now connected by a conducting wire, what is the
amount of charge finally stayed on each sphere?
## Answer:

The charge distribution after the connection is made will be such
that the electric potential on both spheres is the same.
Otherwise charge would transfer along the wire to balance the
potentials at each end of the apparatus. Since the spheres are
far apart compared to their radii, the the charge distribution on
each sphere may be taken as uniform.
You may use Gauss's law to find that the electric field at
the surface of each sphere is
E=k*q/r^{2}, where k is the coulomb constant, q is
the charge on the sphere and r is its radius. The electric
potential on the surface of each sphere is found by integrating
the work done in moving a unit charge in from infinity. This
gives us V=k*q/r for each sphere.

The ratio of the charges then will be
q1/q2=r2/r1=2/1. The total charge is 10-2 or 8 Coulombs.
q1=2q2 and
q1+q2=8. Solve for q1 and q2.

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