Explore BrainMass

# Physics: Electrostatics

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

1. A -3 micro C charge is placed 6 mm away from a -9 micro C charge. What is the potential energy in Joules?

2. What should be the separation in meters, of one charge of +5 micro C and another of +7 micro C so that the repulsion force is equivalent to 8 N?

3. Find the electric field intensity in N/C at distance of 40 cm from the center of a -4 micro C charged sphere that has a radius of 20 cm.

https://brainmass.com/physics/coulombs-law/physics-electrostatics-410836

## SOLUTION This solution is FREE courtesy of BrainMass!

1. Potential energy of a configuration of two charges Q1 and Q2 separated by a distance r is given by: PE = k Q1Q2/r where k = 1/4Пε0 = 9 x 109

Substituting Q1 = - 3 μC, Q2 = -9 μC, r = 6 x 10-3 m we get:

PE = 9 x 109 x (-3 x 10-6)( -9 x 10-6)/ 6 x 10-3 = 40.5 Joule

2. By Coulomb's law force between two charges Q1 and Q2 separated by a distance r is given by: F = k Q1Q2/r2 where k = 1/4Пε0 = 9 x 109

Substituting Q1 = +5 μC, Q2 =7 μC, F = 8 N we get:

8 = 9x109x5x10-6x7x10-6/r2 = 0.315/r2

r2 = 0.315/8 = 0.039 m

r = 0.2 m

3. As the point at which the electric field intensity is to be determined lies out side the sphere and for such a point, the charge on the sphere behaves as if it were concentrated at its center; the electric field intensity is given by:

E = kQ/r2 = k Q/r2 = 9x109x(-4x10-6)/0.42 = - 2.25x105 N/C

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!