Answer:
a. 55.6v
b. [tex]1.61*10^{-7}J[/tex]
Explanation:
Data given
charge 1=+4.10nC
charge 2 =-2.40nC
distance, r= 55cm =0.55m
a. the electric potential at the mid point is the sum of the potential due to individual charge.
The electric potential is expressed as
[tex]V=\frac{kq}{r}\\[/tex]
since we are interested in the electric potential at the mid point, we have
[tex]V=\frac{kq_1}{r/2}-\frac{kq_2}{r/2}\\ V=\frac{2k}{r}(q_1-q_2)\\ V=\frac{2*9*10^9}{0.55}(4.1-2.4)*10^{-9}\\ V=55.6v[/tex]
Hence the electric potential at the mid-point is 55.6v
b. to calculate the potential energy, we use the formula below
[tex]U=\frac{kq_1q_2}{r} \\U=\frac{9*10^9 *4.10*10^{-9}*2.4*10^{-9}}{0.55}\\ U=1.61*10^{-7}J[/tex]
