Answer:
[tex]\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
Explanation:
Electric Potential of Point Charges
The electric potential from a point charge Q at a distance r from the charge is
[tex]\displaystyle V=k\frac{Q}{r}[/tex]
Where k is the Coulomb's constant. The total electric potential for a system of point charges is equal to the scalar sum of their individual potentials. The potential is not a vector, so there is no direction or vectors to deal with.
We are required to compute the total electric potential in the center of the square. We need to know the distance from each corner to the center. The diagonal of the square is
[tex]d=\sqrt2 a[/tex]
where a is the length of the side.
The distance from any corner to the center is half that diagonal, thus
[tex]\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}[/tex]
The total potential in the center is
[tex]V_t=V_1+V_2+V_3+V_4[/tex]
Please note all the potentials must be calculated including the sign of the charges. Since all the charges are equal to Q, and the distances are the same, the total potential is 4 times the individual potential of each charge.
[tex]V_t=4\times V[/tex]
[tex]\displaystyle V=9\times 10^9 \frac{Q}{\frac{a}{\sqrt{2}}}[/tex]
Operating
[tex]\displaystyle V=9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
Thus:
[tex]\displaystyle V_t=4\times 9\sqrt{2}\times 10^9 \frac{Q}{a}[/tex]
[tex]\boxed{\displaystyle V_t=36\sqrt{2}\times 10^9 \frac{Q}{a}}[/tex]
