Question

Please answer part b, thank you!

p03 Problem 17.27 Find the magnitude of the net force on q,-+7μ q,-9,-+7pC 0 N Because the net force acting on q2 is zero there is no direction of the force to find. Find the magnitude of the net force on q,-+7μC q1 +7μC and q3--7μC. Figure 43 x (m) -1 01 2345 6 15

Answer 1

Electrostatic force is given by:

F = k*q1*q2/r^2

Force will be attractive if both charge has opposite sign and Force will be repulsive if both charge has same signs,

Now net force on Charge Q2 will be

F12 (force due to charge Q1 on charge Q2) will be repulsive force and towards the +ve x-axis

F23 (force due to charge Q3 on charge Q2) will be attractive force and towards the +ve x-axis

So, Net force will be

F2 = F12 + F23

So,Magnitude of F2 will be

|F2| = k*|Q1|*|Q2|/r12^2 + k*|Q2|*|Q3|/r23^2

r12 = distance between Q1 and Q2 = 3 - 1 = 2 m

r23 = distance between Q2 and Q3 = 5 - 3 = 2 m

|Q1| = |Q2| = |Q3| = 7 uC

|F2| = 9*10^9*7*10^-6*7*10^-6/2^2 + 9*10^9*7*10^-6*7*10^-6/2^2

|F2| = 0.220 N

Please Upvote.

Comment below if you have any query.