Answer : F = 1.792 N
Explanation :
It is given that,
Magnitude of test charge, [tex]q=6.4\times 10^{-6}\ C[/tex]
Electric field strength, [tex]E=2.8\times 10^5\ N/C[/tex]
We know that the electric field is defined as the force acting per unit electric charge.
[tex]E=\dfrac{F}{q}[/tex]
[tex]F=qE[/tex]
[tex]F=6.4\times 10^{-6}\ C\times 2.8\times 10^5\ N/C[/tex]
[tex]F=1.792\ N[/tex]
Hence, this is the required solution.
