# Exercises On Electric Energy and Potential

See Electric Potential Energy
For Definition And Concepts

## Exercise One.

Show that the work required to assemble four identical point charges of magnitude q on the corners of a square of side a is
5.41 keq2/a.

Solution:
To calculate the work to assemble any number of point charges you can proceed as follows:
- Numerate each charge.
- Assume charges are brought from an infinite distance.
- Charge 1 is brought first and placed on corner 1. This charge does not requires any work because there is no electric field present.
- Charge 2 is brought then and placed on corner 2, charge 3 is brought and placed on corner 3 and so on.

Charge 2 requires a work keq2/a.
Charge 3 requires a work keq2/a + keq2/(a21/2) (work due to charges 1 and 2 already present).
Charge 4 requires a work keq2/a + keq2/(a21/2) + keq2a (work due to charges 1, 2 and 3 already present).

The work needed to get the desired configuration is obtained adding each required work:

keq2/a + keq2/a + keq2/(a21/2) + keq2/a + keq2/(a21/2) + keq2/a. = (keq2/a)(1 + 1 + 2-1/2 + 1 + 2-1/2 + 1) = 5.41 keq2/a.

## Exercise Two

The axis x is the axis of symmetry of a ring uniformly charged of radius R and total charge q. A point charge q with mass m is placed at the center of the ring. When the ring is slightly displaced to the left the point charge moves along the x axis towards infinity. Show that the final speed v of the point charge is v = (2keq2/mR)1/2.

Solution:
The charge potential energy at the center of the ring can be calculated using the relation for the potential in differential form, dV = kedq/R, where dq is an infinitesimal charge on the ring and R is the distance to the center of the ring. The total potential at the center of the ring is the integral of dV, V = keq/R.
Then, the point charge energy is Vq = keq2/R.

At infinity, V = 0 and the energy is zero.
The change in potential energy is 0 - keq2/R = - keq2/R.
The change in kinetic energy is ½ mv2 - 0 = ½ mv2.
As ( Ufinal - Uinitial) + (Kfinal - Kinitial) = 0
- keq2/R + ½ mv2 = 0
½ mv2 = keq2/R or v = (2keq2/mR)1/2.

## Exercise Three

A small sphere of mass 0.2 g hangs by a thread between two parallel vertical plates 5 cm apart. The charge on the sphere is 6x10-9 C. What is the potential difference between the plates if the thread assumes an angle of 10º with the vertical? Assume a uniform electric field between the plates.

Solution:
From the sphere free body diagram, we have:
(A) Tsin10º = F, F = qE, where E is the electric field assumed uniform. Tsin10º = qE.
Also, Tcos10º = mg or T = mg/cos10º =
0.0002 kgx(9.8 m/s2)/cos10º = 0.00199 N
Replacing in (A) and solving for E,
E = Tsin10º/q = 57.6 x103 N/C or 57.6x103 Volts/m.
As in a uniform field E = potential difference (V)/plates separation (d), the potential difference is Ed = (57.6x103 Volts/m)(0.05m) = 2.88 kV.