How Does Charge Distribution Affect Electric Potential at a Point?

[stunner5000pt]

There is a rod with uniform charge distribution lambda . Point P is located directly above the rod at a distance d from the centre of the rod. Find the electric potential at point P

Also in thee second part similar situation however one half of the rod has a negative charge distribution while the other is positivelambda.

For the first part

$$V = 2 \int \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{r}$$ where r is the distance from the poin P to the infinitesimal point on the rod and x is the horizontal axsi upon whihc the rod sits

so then $$r = \sqrt{d^2 + x^2}$$
and then it follows that

$$V = 2 \int \frac{1}{4 \pi \epsilon_{0}} \frac{\lambda dx}{\sqrt{d^2 + x^2}}$$


For the second part

wouldnt i get a zero answer because instead of 2 times the expression you'd get one of those expresions and then the negative of it which when you add them (right?) they cancel out?

 

[wais]

Yes, you are correct. In the second part, the electric potential at point P would be zero because the electric fields from the positive and negative halves of the rod would cancel each other out. This is due to the principle of superposition, which states that the total electric field at a point is the sum of the individual electric fields from each source. Since the electric fields from the positive and negative halves of the rod have equal magnitude but opposite direction, they cancel out and result in a net electric potential of zero.