**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 10
N/C ^{5}

(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 10 N/C) (0.05 m)
= -1.5 x 10^{5}^{4} 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 4**

^{th}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 cm^{2} ]^{1/2} = 5 cm

Then V =

(9
x 10^{9} Nm^{2}C^{-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 (4pr^{2}).

But
E = Q / 4p e_{o} r^{2
}

So;

f = Q / 4p e_{o} r^{2 } x ( 4pr^{2}) = **Q / **

*e*

_{o}_{.}

The charge density s = Q/ A = **Q / ** (4pr

^{2})

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