Answer:
Net electric field, [tex]E_{net}=91406.24\ N/C[/tex]
Explanation:
Given that,
Charge 1, [tex]q_1=7.5\ nC=7.5\times 10^{-9}\ C[/tex]
Charge 2, [tex]q_2=-2.9\ nC=-2.9\times 10^{-9}\ C[/tex]
distance, d = 3.2 cm = 0.032 m
Electric field due to charge 1 is given by :
[tex]E_1=\dfrac{kq_1}{r^2}[/tex]
[tex]E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}[/tex]
[tex]E_1=65917.96\ N/C[/tex]
Electric field due to charge 2 is given by :
[tex]E_2=\dfrac{kq_2}{r^2}[/tex]
[tex]E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}[/tex]
[tex]E_2=25488.28\ N/C[/tex]
The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :
[tex]E_{net}=E_1+E_2[/tex]
[tex]E_{net}=65917.96+25488.28[/tex]
[tex]E_{net}=91406.24\ N/C[/tex]
So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.
