Answer:
V = 50.91 kV
Explanation:
given data
charges = 2.0 µC = 2.0 × [tex]10^{-6}[/tex] C
length of a =2.0 m
solution
we get here first distance by corner to center that is express as
r = a ÷ √2 ...............1
put here value and we get
r = 2.0 ÷ √2
r = √2 m
and
here potential at distance r due the point charge (q) is express as
V = q ÷ ( 4 π εo r ) ...............2
so 4 charges the total potential will be as
V = 4q ÷ ( 4 π εo r ) ...............3
and here 1 ÷ (4 π εo ) = 9.0 × [tex]10^{9}[/tex] N m²/C²
so that
total potential at the center will be
V = ( 4 × 9.0 × [tex]10^{9}[/tex] × 2.0 × [tex]10^{-6}[/tex] ) ÷ √2
solve it and we will get
V = 50.91 kV
The sum of the electric potential at the center of the square is 5.091 x 10⁴V.
Electric potential is given as;
V = kq/r
where;
r is the distance
The diagonal distance from the edge of the square to the center of the square is calculated as follows;
d² = a² + a²
d² = 2a²
d = √(2a²)
d = a√2
d = 2√2
Half of the diagonal length, r = √2
Sum of the electric potential at the center of the square.
[tex]V_{total} = \frac{kq}{\sqrt{2} } \times 4\\\\V = \frac{9 \times 10^9 \times 2\times 10^{-6}}{\sqrt{2} } \times 4\\\\V = 5.091 \times 10^4 \ V[/tex]
The sum of electric potential at infinity is zero.
Learn more about electric potential here: https://brainly.com/question/14306881
