A point charge of 10 μ C is placed at the origin. At what location on the X-axis should a point charge of 40 μ C be placed so that the net electric field is zero at x = 2 cm on the X-axis? (a) x = 6 cm (b) x = 8 cm (c) x = 4 cm (d) x = -4 cm

Answer: (a). Field of 10 at x = 2 cm points along +x, so the 40 charge must be to the right of x = 2 cm. (k(10))/(2²) = (k(40))/(r²) ⇒ r² = 16 ⇒ r = 4 cm Position = 2 + 4 = 6 cm So the answer is (a) x = 6 cm.

A point charge of 10 μ C is placed at the origin. At what location on… | JEE Main 2023 (25 Jan, Shift 2)

Q21JEE Main 2023 · 25 JAN · SHIFT 2Medium

Electric Field & Field Lines

A point charge of 10 $\mu$ C is placed at the origin. At what location on the X-axis should a point charge of 40 $\mu$ C be placed so that the net electric field is zero at $x = 2$ cm on the X-axis?

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Answer: (a)

Field of $10\,\mu\text{C}$ at $x = 2$ cm points along $+x$ , so the $40\,\mu\text{C}$ charge must be to the right of $x = 2$ cm.
$\dfrac{k(10)}{2^2} = \dfrac{k(40)}{r^2} \Rightarrow r^2 = 16 \Rightarrow r = 4$ cm
Position $= 2 + 4 = 6$ cm
So the answer is (a) $x = 6$ cm.

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