Energy and Potential

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Electric Potential Energy

For Definition And Concepts

** 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 k _{e}q^{2}/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 k _{e}q^{2}/a.**

Charge 3 requires a work k_{e}q^{2}/a + k_{e}q^{2}/(a2^{1/2}) (work due to charges 1 and 2 already present).

Charge 4 requires a work k_{e}q^{2}/a + k_{e}q^{2}/(a2^{1/2}) + k_{e}q^{2}a (work due to charges 1, 2 and 3 already present).

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

k_{e}q^{2}/a + k_{e}q^{2}/a + k_{e}q^{2}/(a2^{1/2)} + k_{e}q^{2}/a + k_{e}q^{2}/(a2^{1/2}) + k_{e}q^{2}/a. = (k_{e}q^{2}/a)(1 + 1 + 2^{-1/2} + 1 + 2^{-1/2} + 1) = 5.41 k_{e}q^{2}/a.

** 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 = (2k _{e}q^{2}/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 = k_{e}q/R.

Then, the point charge energy is Vq = k_{e}q^{2}/R.

At infinity, V = 0 and the energy is zero.

The change in potential energy is 0 - k_{e}q^{2}/R = - k_{e}q^{2}/R.

The change in kinetic energy is ½ mv^{2} - 0 = ½ mv^{2}.

As ( Ufinal - Uinitial) + (Kfinal - Kinitial) = 0

- k_{e}q^{2}/R + ½ mv^{2} = 0

½ mv^{2} = k_{e}q^{2}/R or v = (2k_{e}q^{2}/mR)^{1/2}.

** 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/s^{2})/cos10º = 0.00199 N

Replacing in (A) and solving for E,

E = Tsin10º/q = 57.6 x10^{3} N/C or 57.6x10^{3} Volts/m.

As in a uniform field E = potential difference (V)/plates separation (d),
the potential difference is Ed = (57.6x10^{3} Volts/m)(0.05m) = 2.88
kV.

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