How Do You Calculate Electric Potential at a Point Near a Charged Rod?

[Tassadar27]

There is a plastic rod with a length L and a uniform positive charge lying on the x axis. With V = 0 at infinity find the electric potential at point P on the axis at distance d from one end of the rod. I got an answer but I have no clue if it is right or not. This is what I know:

dv = \frac{k dq}{x} \ \ \ \ \ \lambda = \frac{q}{x}

dv = \frac{k\lambda dx}{x} \ \ \ \ \ dq = \lambda dx

\int dv = k\lambda \int \frac{dx}{x} Upper = d + L Lower = d

V = k\lambda(ln(d+L) - ln(d))

Is there anything I am missing?

The next problem is similar but \lambda = cx where c is some positive constant. If I just substitute in cx then the x cancles leaving dx which gives x afer the integral. The answer contains a ln so I am missing something. Any help is appreciated.

 

[Gamma]

Seems right to me. For part ii, V = KCL. Answer does not have an ln.

 

[thepatientmental]

Your approach to finding the electric potential at point P is correct. You have correctly used the formula for electric potential, which is V = kq/r, where k is the Coulomb's constant, q is the charge, and r is the distance from the point to the charge. In this case, you have a uniformly charged rod, so you can use the formula for linear charge density, which is λ = q/L, where q is the total charge and L is the length of the rod.

You have correctly identified that the charge element dq = λdx, where dx is an infinitesimal length of the rod. Then, you have integrated over the length of the rod, from d to d+L, to find the total electric potential at point P. Your final answer, V = kλ(ln(d+L) - ln(d)), is correct.

For the next problem, where λ = cx, you are correct that the x cancels out after substituting in cx for λ. However, you need to be careful when integrating, as the limits of integration will be different. In this case, the limits will be from d to d+L/c, since the charge element, dq = λdx, will now be a function of x. After integrating, you will get V = kcx(ln(d+L/c) - ln(d)).

Make sure you always carefully consider the limits of integration when solving problems involving electric potential.