here the charge density of metal plate is given as

now the electric field is given Gauss law

now here E = constant
so we will have

Since total area on both sides of plate will be double and becomes 2A



Now if we will find the electric field inside the metal plate
Then as we know that total charge inside the plate will always be zero
so we have

Since the acceleration is constant, the average velocity is simply the average of the initial and final velocities of the body:

We can proof that the distance covered by the body moving at constant average velocity
is equal to the distance covered by the body moving at constant acceleration a:
- body moving at constant velocity
: distance is given by

- body moving at constant acceleration
: distance is given by

To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.
The energy fluency is given by the equation

Where
The energy fluency
c = Activity of the source
r = distance
E = electric field
In the other hand we have the equation for current in materials, which is given by

Then replacing our values we have that


We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.
Now the energy fluency would be,



The uncollided flux density at the outer surface of the tank nearest the source is 
Given:
distance from the projector lens to the image, di
projector lens focal length, f
distance from the transparency to the projector lens, do
thin lens equation: 1/f = 1/di + 1/do
do = 4 inches
di = 8 feet
convert feet to inches, for uniformity.
1 foot = 12 inches
8 feet * 12 inches/ft = 96 inches
1/f = 1/96 inches + 1/4 inches
Adding fractions, denominator must be the same.
1/f = (1/96 * 1/1) + (1/4 * 24/24)
1/f = 1/96 + 24/96
1/f = 25/96
to find the value of f, do cross multiplication
1*96 = f * 25
96 = 25f
96/25 = f
3.84 = f
The focal length of the project lens is 3.84 inches