Question

Two very large, nonconducting plastic sheets, each 10.0 cm thick, carry uniform charge densities σ1,σ2,σ3and σ4 on their surfaces, as shown in the following figure(Figure 1). These surface charge densities have the values σ1 = -5.50 μC/m2 , σ2=5.00μC/m2, σ3= 1.60 μC/m2 , and σ4=4.00μC/m2. Use Gauss's law to find the magnitude and direction of the electric field at the following points, far from the edges of these sheets.

What is the magnitude of the electric field at point A, 5.00 cm from the left face of the left-hand sheet?

Express your answer to three significant figures and include the appropriate units.

What is the direction of the electric field at point A, 5.00 cm from the left face of the left-hand sheet?

What is the direction of the electric field at point , 5.00  from the left face of the left-hand sheet?

to the left.

to the right.

upwards.

downwards.

What is the magnitude of the electric field at point B, 1.25 cm from the inner surface of the right-hand sheet?

Express your answer to three significant figures and include the appropriate units.

What is the direction of the electric field atpoint B, 1.25 cm from the inner surface of the right-hand sheet?

What is the direction of the electric field atpoint B , 1.25  from the inner surface of the right-hand sheet?

to the left.

to the right.

upwards.

downwards.

What is the magnitude of the electric field at point C, in the middle of the right-hand sheet?

Express your answer to three significant figures and include the appropriate units.

What is the direction of the electric field at point C, in the middle of the right-hand sheet?

to the left.

to the right.

upwards.

downwards.

Answer 1

We know that the electric field due to a plane carrying a uniform surface charge is where is the charge density and is a unit vector pointing away from the surface.

a) At A the electric field due to the left face of the left sheet= pointing to the right.

The field due to the right face of the left sheet= to the left.

The field due to the left face of the right sheet= to the left.

The field due to the right face of the right sheet= to the left.

Therefore the net field at A=

to the left.

b)At B:

The electric field due to the left face of the left sheet= to the left.

The field due to the right face of the left sheet= to the right.

The field due to the left face of the right sheet= to the left.

The field due to the right face of the right sheet= to the left.

The net field=

c) At C:
The field due to the left face of the left sheet= to the left.

The field due to the right face of the left sheet= to the right.

The field due to the left face of the right sheet= to the right.

The field due to the right face of the right sheet= to the left.

The net field=

to the left.

Note:.