Question

Consider an equilateral triangle with sides measuring 1.0 m in length. At each point of the triangle is a +2.0 μC charge. Calculate the Coulomb force on each charge. Recall that forces are vectors and thus your answer will require a magnitude and direction for each of the three forces.

Answer 1

Th magnitude of each Force will be
`=62.35*10^(-3)\ \rm N`

Step-by-step explanation:

Given:

• Length of each side of the equilateral triangle, L=1 m
• Magnitude of each point charge
  `Q=2\ \rm \mu C`

Since all the charges are identical and distance between them is same so magnitude of the force between each charge is equal.

Let F be the force between the particles. According to Coulombs Law we have

`F=(kQ^2)/(L^2)\\=(9*10^9* (2*10^(-6))^2)/(1^2)\\F=36*10^(-3)\ \rm N`

Now the the force on any charge by other two charges will be F and the angle between the two force is
`60^\circ`

Let
`F_(resultant)` be the force on nay charge by other two

By using vector Law of addition we have

`F_(resultant)=√((F^2+F^2+2F* F * cos60^\circ))\\=√(3)F\\=√(3)*36*10^(-3)\ \rm N\\=62.35*10^(-3)\ \rm N`

The angle made by the resultant vector will be

`\tan\beta=(F\sin60^\circ)/(F+F\cos60^\circ)\\\beta=30^\circ`