A charge of 3 C is at (3, -1 ) and a charge of -4 C is at ( -3, 7 ). If both coordinates are in meters, what is the force between the charges?

1 Answer
Jul 6, 2018

F=-1.08*10^9 N

Explanation:

Charge q_1=3C is placed at (3,-1)
Charge q_2=-4C is placed at (-3,7)

Distance between both charges is
r=sqrt((x_2-x_1)^2+(y_2-y_1)^2
r=sqrt((-3-(3))^2+(7-(-1))^2
r=sqrt((-6)^2+(8)^2
r=sqrt(36+64)
r=sqrt(100) m = 10 m and r^2=100

Force between two charges q_1 and q_2 is
F=(kq_1q_2)/r^2 where k=9*10^9 Nm^2/C^2
=>F=((9*10^9)(3)(-4))/(100)
=>F=((9*10^9)(-12))/(100)
=>F=((-108*10^9))/(100) N
=>F=-1.08*10^9 N