homework helper, contact and send problems

homework by email, send
helper, math and physics
math tutor web site
physics help web site
math and physics help, fees
help contact
problems, send here
forward payments here
teacher background

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.

Related Sites: · Physics, Main Page
· Physics, Mathematics
· Physics, Detailed Homework Scope Help
· Energy, Work and Power: Concepts
· Kinetic Energy
· Potential Energy
· Power
· Physics Problems, Example
· Physics Homework - Mechanical Energy Conservation Problems
· Physics Homework - Mechanical Power Problems
· Coulomb's Law
· Exercises Using Coulomb's Law
· Electric Field Charges
· Electric Field Exercises
· Ohm's Law, Principle
· Ohm's Law Exercises
· Gauss' Law
· Gauss' Law Exercises
· Second Newton's Law
· Second Newton's Law Examples, Part One
· Second Newton's Law Examples, Part Two
· Sound Waves
· Sound Waves: Standing, Interference, Doppler Effect - Examples
· Sound Waves, Doppler Effect - Examples
· Vectors, Scalars
· Vectors, Scalars - Analytic Method
· Addition Vector Tools, Problems
· Free Fall Theory
· Free Fall Exercises, Part One
· Free Fall Exercises, Part Two
· Free Fall Exercises, Part Three

Bookmark and Share