Answer:
The force on the test charge is 114.75 N.
Explanation:
Given that,
Test charge [tex]q=6\ \mu C[/tex]
Suppose the test charge is 3.4 μC and the distance between the charge and test charge is 4.0 cm.
We need to calculate the force on the test charge
Using formula of electric force
[tex]F=\dfrac{kqq_{1}}{r^2}[/tex]
Where, q = test charge
q₁ =charge
r = distance
Put the value into the formula
[tex]F=\dfrac{9\times10^{9}\times6\times10^{-6}\times3.4\times10^{-6}}{(4.0\times10^{-2})^2}[/tex]
[tex]F=114.75\ N[/tex]
Hence, The force on the test charge is 114.75 N.