Answer:
- 91499.95 V
Explanation:
Let the mid point is P.
qA = 3.4 pC = 3.4 x 10^-12 C
qB = - 6.10 micro Coulomb = - 6.10 x 10^-6 C
AB = 1.20 m
AP = BP = 0.6 m
Let the potential at P due to the charge at A is VA and the potential at P due to the charge at B is VB.
The potential at P is V
V = VA + VB
[tex]V = \frac{Kq_{A}}{AP}+\frac{Kq_{B}}{BP}[/tex]
[tex]V = \frac{9\times 10^{9}\times 3.4\times 10^{-12}}{0.6}-\frac{9\times 10^{9}\times 6.10\times 10^{-6}}{0.6}[/tex]
V = 0.051 - 91500
V = - 91499.95 V