Answer:
d. 0V
Explanation:
The magnitude of four point charges are +3q, -q, +2q and -4q. I think you forget to mention the signs.
As we know that the potential due to the point charge that has traveled the distance d can be represented mathematically as,
[tex]V = k\frac{q}{r}[/tex]
[tex]k[/tex] = 1/4λε = 9×[tex]10^{9}[/tex] Nm²/C²
Now as it is mentioned in the question that all four charges are arranged in the corners of a square so there distance from the center is same. We can rewrite the above potential equation as follows.
[tex]V = \frac{k}{d} (q_{1} + q_{2} +q_{3} +q_{4} )[/tex] (1)
We can find out d by the pythagoras theorem, as we are dealing with square so d is a semi diagonal.
[tex]d = \sqrt{\frac{L^{2} }{4}+\frac{L^{2} }{4} } = \frac{\sqrt{2} }{2} L[/tex]
by putting all values in equation (1)
V = [tex]\frac{9*10^{9} }{\frac{\sqrt{2} }{2} L } (+3q-q+2q-4q)[/tex]
V = 0V
