Answer:
V = 50.91 kV
Explanation:
given data
charges = 2.0 µC = 2.0 ×
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 ×
N m²/C²
so that
total potential at the center will be
V = ( 4 × 9.0 ×
× 2.0 ×
) ÷ √2
solve it and we will get
V = 50.91 kV