Explanation:
(1) Given that,
A charge, [tex]q=3\times 10^{-7}\ C[/tex]
Electric field, E = 27 N/C
We need to find the force acting on the charge. The force on the charge is given by :
[tex]F=qE\\\\F=3\times 10^{-7}\times 27\\F=8.1\times 10^{-6}\ N[/tex]
So, the force acting on the charge is [tex]8.1\times 10^{-6}\ N[/tex]
(b) Charge, [tex]q=5\times 10^{-6}\ C[/tex]
Force, [tex]F=2\times 10^{-4}\ N[/tex]
Let E be the electric field.
[tex]E=\dfrac{F}{q}\\\\E=\dfrac{2\times 10^{-4}}{5\times 10^{-6}}\\\\E=40\ N/C[/tex]
So, the electric field is 40 N/C.
