Question

Three point charges are arranged in a horizontal line as shown below. Find the electric forces (in units of kQ²/R²) on the charges given the following: Q₁ = 100 Q, Q₂ = 25 Q, Q₃ = 300 Q, r₁ = 5 R, and r₂ = 5 R. Remember that a positive force points to the right and a negative force points to the left.

$charges_3_line.jpg$

What is the net force on charge Q₁?

What is the net force on charge Q₂?

What is the net force on charge Q₃?

What is the sum of the forces on all three charges?

I am thinking that when I draw my free body diagram I should get the following postive or negative values:

F31 is going to be to the RIGHT so positive value

F32 is going to be to the RIGHT so positive value

F12 is going to be to the left so negative value

F21 is going to be to the left so negative value.

F23 is going to be to the RIGHT so positive value.

Answer 1

The three charges Q_{1 ,} Q₂ , Q₃ are on a horizontal line staring from left to right ,  Q₁ , Q_{2 are seperated by 5R}

_and Q₂ , Q_{3 are seperated by 5R.}

_{Net force on charge} Q₁ is $F_{12}+F_{13}=-\frac{k Q_1Q_2}{(5R)^2} - \frac{k Q_1Q_3}{(10R)^2}=\frac{-kQ^2}{R^2}\times400\, N$

_{Net force on charge} Q₂ is $F_{21}+F_{23}=\frac{k Q_2Q_3}{(5R)^2} - \frac{k Q_2Q_3}{(5R)^2}=\frac{-kQ^2}{R^2}\times200\, N$

_{Net force on charge} Q₃  is $F_{31}+F_{32}=\frac{k Q_3Q_1}{(10R)^2} + \frac{k Q_3Q_2}{(5R)^2}=\frac{kQ^2}{R^2}\times600\, N$