Question

1.12 [2pt] Consider one infinitely long straight wire with a uniform charge density of 1 C/m. Sketch the electric field around the wire

1.12 ANSWER

1.13 [2pt] In problem 1.12, calculate the magnitude of electric field at a distance R from the wire. How is it different (if any) from the field of a point charge?

1.13 ANSWER

1.14 [2pt] Consider two infinite wires 1 m apart with a uniform charge density per unit length 1 C/m. Calculate the force per unit length between the wires. To do this, just select a tiny, nearly point-like section of a wire of length Land calculate the force due to the electric field of the other wire using the Coulomb’s law. Your answer should be proportional to L. Divide it by L and this is the force per unit length.

1.14 ANSWER

1.15 [1pt] Explain, in a few sentences, how the electric force between two charged wires differs from the regular Coulomb law and why.

1.15 ANSWER

Answer 1

1.12.

Here, the line has a charge density or uniform charge density () of 1C/m. The field lines will look like the adjacent figure.

1.13.

The electric field of an infinite line charge with a uniform linear charge density can be obtained by a using Gauss' law in electrostatics. Considering a Gaussian surface in the form of a cylinder at radius R and length L, the electric field (E) has the same magnitude at every point of the cylinder and is directed outward. The electric flux () is then just the electric field times the area (A) of the cylinder.

If we consider a point charge (Q), which is equivalent to the net charges included inside the Gaussian surface stated above, will be

Therefore, for R = L/2 the electric field will be same if you place the included charge at a point with distance R.

1.14.

Electric field at any point on the line due to the other line will be

The force per unit length between the wires will be

1.15.

The force is inversely proportional to the separation of line charge. But in case of point charge, the force will be inversely proportional to the square of the separation. As discussed in 1.13, you can estimate the force difference between the two cases.

If we consider a point charge (Q), which is equivalent to the net charges included inside the Gaussian surface stated above, will be

Therefore, for R = L/2 the electric field will be same if you place the included charge at a point with distance R.