Athin, uniformly charged insulating rod has a linear charge density λ = 3 nc/m and lies along the x axis from x = 1m to x = 3m.
a) find the electric potential at point a that has the coordinates (0,4).
b) find the electric field at that location.

A)

B) The vector definition of E-field is

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

By using an integral table, this can be calculated:

B) The electric field can be found by a similar approach, but a different formula:

Let's apply this formula to the infinitesimal part we have chosen.

By the geometry sine and cosine terms can be found:

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

So, the final E-field is

The magnitude of the E-field is

E = 2.66 N/m

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.