Answer:
The total potential (magnitude only) is 11045.45 V
Explanation:
Electric Potential
The total electric potential at location A is the sum of all four individual potentials produced by the charges, including the sign since the potential is a scalar magnitude that can be computed by
[tex]\displaystyle V=\frac{kq}{r}[/tex]
Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:
[tex]\displaystyle V_1=\frac{9\cdot 10^{9}\times -2.16\cdot 10^{-6}}{0.88}=-22090.91\ V[/tex]
The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical
[tex]V_2=-22090.91\ V[/tex]
The potential of the topmost charge is almost equal to the above computed, is only different in the sign:
[tex]V_3=+22090.91\ V[/tex]
The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':
[tex]\displaystyle V_4=\frac{9\cdot 10^{9}\times 2.16\cdot 10^{-6}}{1.76}=+11045.45 \ V[/tex]
The total electric potential in A is
[tex]V=-22090.91\ V-22090.91\ V+22090.91\ V+11045.45 \ V[/tex]
[tex]V=-11045.45 \ V[/tex]
The total potential (magnitude only) is 11045.45 V
