Question

Two positive charges q₁ and q₂ lie along a straight line separated by a distance of 2 m as shown. 

(a) Find a location along the straight line joining the two charges, where if a positive charge q₃ is placed, it experiences a zero-resultant force. 

(b) Will the resultant force on q₃ placed at the location of part (a) still be zero, if it is negatively charged? Explain.

Answer 1

(a) For a resultant force at the location of q₃ to be zero, the net electrostatic force on q₃ due to q₁ and q₂ has to be zero. 

Since q₃ is positive, it will be under the effect of repulsive force by both q₁ and q₂ as represented by F₁₃ and F₂₃. 

That is,

\(F_{13} = F_{23}\)

\( \frac{kq_1q_3}{x^2} = \frac{kq_2q_3}{(2-x)^2}\)

Substitute for values of q₁ = 2C, q₂ = 4C and solve to get,

\((2-x)^2 = 2x^2\)

Solve for x,

\(x= \frac{2}{\sqrt{2} + 1} = 0.83m\) 

So q₃ placed at 0.83 m away from q₁ along the straight line joining q₁ and q₂ experiences a zero resultant force. 

• 1 mark for drawing the correct diagram and explanation of forces acting on charge q₃ 
• 1 mark for writing a correct equation for forces on q₃ using Coulombs law 
• 1 mark for substituting and solving for the value of x 

(b) Yes, the negative charge at the location of q₃ will experience zero resultant force. 

The forces on the negative charge due to q₁ and q₂ will get reversed.