A very large plastic slab has a finite thickness of 2.60 cm. Take the slab to be

Question

A very large plastic slab has a finite thickness of 2.60 cm. Take the slab to be infinite in the x and y directions with its central plane at z = 0, so that the surfaces of the slab are at z = ?1.30cm and z = +1.30 cm.

(a) Assume the plastic has a uniform volume charge density of 290.0 ?C/m3.
With the given volume charge density, what is the charge per unit area on the planar slab?
C/m2

Find the electric field strength at the following values of z. HINT: Begin by solving the problem symbolically, finding expressions for the electric field strength both inside and outside the slab.

at z = 0:  
at z = 0.650 cm:  N/C

at z = 1.30 cm:  N/C

at z = 2.60 cm:  N/C

(b) Instead of the situation in (a), assume the plastic has a nonuniform volume charge density given by Cz2, where C is a constant given by 3230.0 ?C/m5.
With this nonuniform volume charge density, what is the charge per unit area on the planar slab?
C/m2

Find the electric field strength at the following values of z. HINT: Begin by solving the problem symbolically, finding expressions for the electric field strength both inside and outside the slab.

at z = 0:  

at z = 0.650 cm:  N/C

at z = 1.30 cm:  N/C

at z = 2.60 cm:  N/C

Explanation / Answer

to find the field strength midway between the charges you must assume there is a positive test charge present.

im assuming that by -4nC you mean -4 "nanocoulombs" (1 x 10^-9). not "microcoulombs" (1 x 10^-6)

convert to coulombs:
-4nC = -4 x 10^-9 C
-6nC = -6 x 10^-9 C

you should know:

E = (kQ)/r^2
k = 9 x 10^9 {its just general a constant}
Q = magnitude of charge (units: coulombs C)
r = radius or distance from charge (units: m)

Also, you should be able to add simple 2D vectors.

Calculations

I'm assuming that if you drew a diagram the -4nC would be on the left and the -6nC on the right, the midpoint charge located on the direct line between them

-4nC -------- x -------- -6nC

magnitude of E between -4nC and point charge

E = (kQ)/r^2
E = ((9 x 10^9)(4 x 10^-9)) / ((0.2)^2)
E = 900 N/C to the right

magnitude of E between -6nC and point charge

E = (kQ)/r^2
E = ((9 x 10^9)(6 x 10^-9)) / ((0.2)^2)
E = 1350 N/C to the left

Let "left" be the positive direction

E(total) = 1350 - 900 {consider positive direction to be to the left, therefore making 1350N/C to the left positive and 900N/C to the right negative (i.e. -900 N/C to the left)}
= 450 N/C to the left

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