Manipal MET2018PhysicsElectrostatics
The distance between charges 5 10⁻¹¹ C and -2.7 10⁻¹¹ C is 0.2 m . The distance at which is third charge should be placed from 4 e in order that it will not experience any force along the line joining the two charges is
Options
- A0.44 m
- B0.65 m
- C0.556 m
- D0.350 m
Correct answer
C. 0.556 m
Step-by-step solution
From the question F₁=F₂5 10⁻¹¹ C -2.7 10⁻¹¹ C aligned & 1 4 ₀ 5 10⁻¹¹ q (0.2+x)^2 = 1 4 ₀ 2.7 10⁻¹¹ q x^2 & x=0.556 ~m from IInd charge. aligned