Answer:
The magnitude of net electric is [tex]5\times 10^6\ \rm N/C[/tex] from charge [tex]q_1[/tex] to [tex]q_2[/tex] along the line joining them.
Explanation:
Given:
Distance between the charges =30 cm
Magnitude of charge [tex]q_0 =3\ \rm \mu C[/tex]
Force exerted by [tex]q_1[/tex] on [tex]q_0 [/tex]=25 N
Force exerted by [tex]q_2[/tex] on [tex]q_0 [/tex]=10 N
Now according to coulombs Law we have
[tex]\dfrac{kq_1q_0}{0.08^2}=25\\\\q_1=59.25\times10^{-7}\ \rm C[/tex]
similarly
[tex]\dfrac{kq_2q_0}{0.22^2}=10\\\\q_2=179.25\times10^{-7}\ \rm C[/tex]
Noe the electric field at eh position of charge [tex]q_0[/tex] is given by
Let [tex]E_1[/tex] be the electric field due to charge [tex]q_1`[/tex] and [tex]E_1[/tex] bet he electric field due to charge [tex]q_2`[/tex]
then The net electric Field at the point is given by
[tex]E_{net}=E_1-E_2\\=\dfrac{kq_1}{0.08^2}-\dfrac{kq_2}{0.22^2}\\\\=\dfrac{9\times10^9\times59.25\times10^{-7}}{0.08^2}-\dfrac{9\times 10^9\times 179.25\times 10^{-7}}{0.22^2}\\\\=5\times 10^6\rm N/C[/tex]
The direction of electric Field is from charge [tex]q_1[/tex] to [tex]q_2[/tex] along the line joining them.
