ExamBro
ExamBro
JEE Mains · Physics · STD 12 - 1. Electric charges and fields

A point charge +q is placed at the origin. A second point charge +9 q is placed at \((\mathrm{d}, 0,0)\) in Cartesian coordinate system. The point in between them where the electric field vanishes is :

  1. A \((4 \mathrm{~d} / 3,0,0)\)
  2. B \((\mathrm{d} / 4,0,0)\)
  3. C \((3 \mathrm{~d} / 4,0,0)\)
  4. D \((\mathrm{d} / 3,0,0)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \((\mathrm{d} / 4,0,0)\)

Step-by-step Solution

Detailed explanation

Let \(\mathrm{E}_{\mathrm{p}}=0\) \(\begin{aligned} & \therefore \frac{\mathrm{kq}}{\mathrm{x}^2}=\frac{\mathrm{k} 9 \mathrm{q}}{(\mathrm{~d}-\mathrm{x})^2} \\ & \Rightarrow \frac{\mathrm{~d}-\mathrm{x}}{\mathrm{x}}=3 \Rightarrow \mathrm{x}=\frac{\mathrm{d}}{4} \end{aligned}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app