Please help with calculating force on particle.

Question

T‎wo fix‎ed par‎ticles, ea‎ch of ch‎arge 5.0x10^-6, are 24 cm a‎part. wha‎t for‎ce do th‎ey exe‎rt on thir‎d pa‎rticle of c‎harge -2.5x10^-6 t‎‎hat is 13 cm from eac‎h of th‎em

Express answer in vector form: magnitude + direction.

Answer 1

Please draw a picture of the situation.

Pick one of the positive charges and call its location P.

Let M be the midpoint between the two positive charges.

Let N be the position of the negative charge.

The triangle MNP is a right triangle with hypothenuse NP = 13cm = 0.13 m, and

one side MP 12 cm.

By Pythagorean theorem the other side MN is 5 cm long.

Let α be the angle at N.

By definition of trigonometric functions

cos(α) = MN/NP = 5/13

The positive charge at P exerts an attractive force F on the negative charge at N.

The force F is directed from N towards P.

Equally large force is directed from the negative charge towards the other positive charge.

The net force acting on the negative charge is then the vector sum of these two forces.

The force from N to P can be decomposed into the sum of two forces --

one along MN and the other parallel to MP.

Their magnitudes are Fcos(α) and Fsin(α) respectively.

The other force can be decomposed in a symmetrical way.

The components parallel to MP cancel each other, but the components from N to M add up.

Therefore the net force is directed from N to M and is of magnitude 2Fcos(α).

It only remains to calculate F from Coulomb's law:

F = kq₁q₂/r²

= 9×10⁹×5×10^-6×2.5×10^-6/0.13² ≈ 6.66 N

Therefore the net force is

2Fcos(α) = 2×6.66×5/13 ≈ 5.1 N