1) A particle of charge +3 x 10-9 C is in a uniform E-field directed to the left:
¬E ----------------------x q--------
It is released from rest and moves a distance of 5 cm, after which its kinetic energy is found to be 4.5 x 10-5 J.
(a)What is the magnitude of the electric field?
(b)What is the potential of the starting point relative to the endpoint?
(a) Work is force x displacement, or W = F · s so:
F = W/s = (4.5 x 10-5 J)/ 0.05 m = 9 x 10-4 N and:
E = F/q
= (9 x 10-4 N )/ (3 x
10-9 C) = 3 x 10
(Note here the work done is just the kinetic energy. Why?)
(b) The potential in this case, is V = - E (s), since E = - V/s
V = - (3 x 10
5 N/C) (0.05 m)
= -1.5 x 104 V
2. 1) A charge of (-20 mmC) is placed at one corner of a 3 x 4 cm rectangle while charges of 10 mmC are placed at two adjacent corners as shown:
First obtain the diagonal distance from the (-20 mmC) charge in the upper right corner to the opposite corner. From Pythagoras::
D = [3 cm 2 + 4 cm2 ]1/2 = 5 cm
Then V =
(9 x 109 Nm2C-2)[ 10mmC/ 0.04m + 10mmC/ 0.03m + -20mmC/ 0.045m]
And note all the charges are in pico-Coulombs (10-12 C) which leads to the result: V = 1.65 V or 1.65 J/C.
3) A hollow sphere has a charge of +Q at its center. Find the electric flux from its surface and the charge density.
The flux through the surface is defined:
f = E x (Area) = E x (4pr2).
But E = Q / 4p eo r2
f = Q / 4p eo r2 x ( 4pr2) = Q / eo.
The charge density s = Q/ A = Q / (4pr2)