## Monday, June 21, 2021

### Solutions To Electrostatics Problems (Part 1)

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?

Solution:

(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 105 N/C

(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

So: V = - (3 x 105 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:

Using this data, calculate the potential at the 4th corner. (Note: potential depends only on distances, not directions!)

Solution:

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.

Solution:

The flux through the surface is defined:

f =   E x (Area)  =  E x   (4pr2).

But E =  Q / 4p eo  r2

So;

f =  Q / 4p eo  r   x  (  4pr2) = Q / eo.

The charge density s   =  Q/ A  =    (4pr2)