Charges on Conductors and Gauss's Law

[crazynut52]

Can someone give me a couple hints to get me started on this problem, I am not sure where to go next.

Infinite line of charge has charge per unit unit length 4.8*10^-6 and lies along the x axis. A second uniform line of charge with charge/unit length
-2.4*10^-6 is parallel to the x-axis at y=4. What is the net electric field at y=2 and y=6?

I used the formula E= lamda/2(pi)(epsilon naught)(r) but I am not sure what to put in for r and where to go next.

 

[Tide]

You calculated the electric field produced by the individual line charges so all you have to do is add them to find the total field. You just have to remember that the "r" in each expression represents the distance from a point in space to the line charge AND the direction of the field due to each line charge is given by the displacement VECTOR from the point in space to the line charge.

It's worth noting that there is no x-component to any of these vectors.

 

[wais]

To solve this problem, you can use Gauss's Law, which states that the net electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space (epsilon naught). In this case, we can consider a Gaussian surface in the shape of a cylinder with its axis along the x-axis, passing through the point at y=2 and y=6.

For the point at y=2, the Gaussian surface will enclose both the positive and negative charges, therefore the net charge enclosed will be the difference between the two charges: (4.8*10^-6) - (-2.4*10^-6) = 7.2*10^-6.

For the point at y=6, the Gaussian surface will only enclose the negative charge, so the net charge enclosed will be just -2.4*10^-6.

Now, we can use Gauss's Law to calculate the electric field at these points. The electric flux through the Gaussian surface is given by E*A, where A is the surface area of the cylinder. At y=2, the surface area is 2*pi*r, where r is the distance from the x-axis to the cylinder, which is just 2 units. So the electric flux is 2*pi*r*E.

Setting this equal to the net charge enclosed divided by epsilon naught, we get:

2*pi*r*E = (7.2*10^-6)/epsilon naught

Solving for E, we get:

E = (7.2*10^-6)/(2*pi*r*epsilon naught)

Plugging in r=2 and epsilon naught=8.85*10^-12, we get:

E = 1.022*10^-3 N/C

Similarly, at y=6, the electric flux is just pi*r*E, and setting it equal to the net charge enclosed divided by epsilon naught, we get:

pi*r*E = (-2.4*10^-6)/epsilon naught

Solving for E, we get:

E = (-2.4*10^-6)/(pi*r*epsilon naught)

Plugging in r=6 and epsilon naught=8.85*10^-12, we get:

E = -0.022*10^-3 N/C

Therefore, the net electric field at y=