Answer: 9. 9 times force
8. 7000 N/C directed towards q₂
Explanation:
9.
Let the two charges be [tex]q_1[/tex] and [tex]q_2[/tex] separated by distance r. Then the force of repulsion is given as :
[tex]force= k\frac{q_1 q_2}{r^2}[/tex]
where, k is the Coulomb's constant.
When the magnitude of the charges is tripled: [tex]q_1'=3q_1[/tex] and [tex]q_2'=3q_2[/tex], the new force of repulsion, force' is:
[tex]force'= k \frac{q_1' q_2'}{r^2} = k\frac{3q_1 \times 3q_2}{r^2}= 9k\frac{q_1 q_2}{r^2}=9\times force[/tex]
8.
The electric field is given by:
[tex]E = k\frac{q}{r^2}[/tex]
where q is the charge and r is the distance of the point from the charge where electric field is required to be found.
It is given that, [tex]q_1 = 30nC = 30\times 10^{-9}C[/tex]
[tex]q_2=-45nC= -45\times10^{-9}C[/tex]
Electric field due to charge [tex]q_1[/tex] at point a is:
[tex]E_1=8.99\times10^9N.m^2/C^2\times \frac{30\times 10^{-9}C}{(0.30m)^2} = 2996.67 N/C \text{ directed to } q_2 [/tex]
Electric field due to charge [tex]q_2[/tex] at point a is:
[tex]E_2=8.99\times10^9N.m^2/C^2\times \frac{-45\times 10^{-9}C}{(0.20m)^2} =- 10113.75 N/C\text{ directed to } q_2[/tex]
Thus, net electric field due to both charges at a is
[tex]E=E_1+E_2 = 2996.67+(-10113.75) = -7117.08 N/C = 7000 N/C \text{ directed towards } q_2[/tex]
