Answer:
The electric field intensity of electric force is [tex]1.35\times10^{7}\ N/C[/tex]
Explanation:
Given that,
First charge [tex]q_{1}=15\times10^{-6}\ C[/tex]
Second charge [tex]q_{2}=150\times10^{-6}\ C[/tex]
Distance = 0.1 m
We need to calculate the electric field intensity of electric force
Using formula of intensity of electric force
[tex]E=\dfrac{F}{q_{2}}[/tex]
[tex]E=\dfrac{kq_{1}}{r^2}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times15\times10^{-6}}{(0.1)^2}[/tex]
[tex]E=13500000\ N/C[/tex]
[tex]E=1.35\times10^{7}\ N/C[/tex]
Hence, The electric field intensity of electric force is [tex]1.35\times10^{7}\ N/C[/tex]
