76. We denote q = 25

× 10

−9

C, y = 0.6 m, x = 0.8 m, with V = the net potential (assuming V

→ 0 as

r

→ ∞). Then,

V

A

=

1

4πε

0

q

y

+

1

4πε

0

(

−q)

x

V

B

=

1

4πε

0

q

x

+

1

4πε

0

(

−q)

y

leads to

V

B

− V

A

=

2

4πε

0

q

x

−

2

4πε

0

q

y

=

q

2πε

0

1

x

−

1

y



which yields ∆V =

−187 ≈ −190 V.