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.

• A. 2 cm from +3µC
• B. 3 cm from -2µC
• C. 10 cm from +3µC
• D. No such point exists

Hint:

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.

Answer 1

The Correct Option is A

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: \(\boxed{\text{A}}\)