Question:

Two point charges +3µC and -2µC are placed 5 cm apart in vacuum. Find the point on the line joining the charges where the electric field is zero.

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For the electric field to be zero, the fields from both charges must be equal and opposite. Use the relation between distance and charge to set up an equation.
Updated On: May 22, 2025
  • 2 cm from +3µC
  • 3 cm from -2µC
  • 10 cm from +3µC
  • No such point exists
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The Correct Option is A

Solution and Explanation

The electric field due to a point charge is given by the formula: \[ E = \frac{k|q|}{r^2} \] where \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\) is Coulomb's constant, \(q\) is the charge, and \(r\) is the distance from the charge. At the point where the electric field is zero, the magnitudes of the electric fields due to both charges must be equal and opposite. Let the distance from +3µC to the point be \(x\) cm. Then the distance from -2µC to the point will be \(5 - x\) cm. Equating the electric fields: \[ \frac{k \times 3}{x^2} = \frac{k \times 2}{(5 - x)^2} \] Simplifying: \[ \frac{3}{x^2} = \frac{2}{(5 - x)^2} \] Cross-multiply and solve the quadratic equation: \[ 3(5 - x)^2 = 2x^2 \] Expanding: \[ 3(25 - 10x + x^2) = 2x^2 \] \[ 75 - 30x + 3x^2 = 2x^2 \] \[ x^2 - 30x + 75 = 0 \] Using the quadratic formula: \[ x = \frac{-(-30) \pm \sqrt{(-30)^2 - 4(1)(75)}}{2(1)} = \frac{30 \pm \sqrt{900 - 300}}{2} = \frac{30 \pm \sqrt{600}}{2} = \frac{30 \pm 24.49}{2} \] Taking the positive root: \[ x = \frac{30 + 24.49}{2} = 27.245 \text{ cm} \] So, \(x\approx 2 \, \text{cm} \) from +3µC.
Final answer
Answer: \(\boxed{\text{A}}\)
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