Question

Three positive charges are arranged as shown in the diagram.

What is the magnitude and direction of the net electrical force acting on the +3 µC charge?

magnitude: 1.75 N
direction: 6.2° north of east
magnitude: 1.75 N
direction: 6.2° south of east
magnitude: 1.86 N
direction: 21° north of east
magnitude: 1.86 N
direction: 21° south of east

Answer 1

magnitude: 1.75 N
direction: 6.2° north of east

Answer 2

Answer: magnitude: 1.75 N direction: 6.2° north of east

The Coulomb force equation is,
`F_(12) = (k q_1 q_2)/(r^2)` where,
`F_(12)` is force on 1 due to 2, k is coulomb constant,
`q_1 ,q_2` are the charges and r is the distance between the charges 1 and 2.

The magnitude of force on 3 from 5 is,
`F_(35) = (9 * 10^9 N m^2/C^2 * 3 \mu C (10^(-6)C)/(1 \mu C) * 5 \mu C(10^(-6)C)/(1 \mu C))/((0.45 m)^2)=0.67 N`

The magnitude of force on 3 from 9 is,
`F_(39) = (9 * 10^9 N m^2/C^2 * 3 \mu C (10^(-6)C)/(1 \mu C) * 9 \mu C(10^(-6)C)/(1 \mu C))/((0.45 m)^2)=1.20 N`

The resultant magnitude is,
`F_R=√(A^2+B^2+2AB cos \phi)=\sqrt{F_(35)^2+F_(39)^2+2F_(35)F_(39) cos \phi}`
`=1.75N`

The Force from 9 on 3 is large which is in northeast direction so the resultant force will also be in north east direction only. So, 6.2 ° North of east.