Three point charges q, -2q and 2q are placed on x -axis at a distance x = 0, x = 3/4R and x = R respectively from origin as shown. If q = 2 × 10⁻⁶ C and R = 2 cm, the magnitude of net force experienced by the charge -2q is ___________ N. Numerical answer type. Work it out, then open the answer.

Answer: 5440. On -2q (at 3/4R): q at distance 3/4R attracts it towards -x: (k(q)(2q))/((3R/4)²) = 32/9(kq²)/(R²). 2q at distance R/4 attracts it towards +x: (k(2q)(2q))/((R/4)²) = 64(kq²)/(R²). Net = (64 - 32/9)(kq²)/(R²) = 544/9(kq²)/(R²) (kq²)/(R²) = 9 × 10⁹ × 4 × 10⁻¹²4 × 10⁻⁴ = 90 N Net force = 544/9 × 90 = 5440 N.

Three point charges q, -2q and 2q are placed on x -axis at a distance… | JEE Main 2023 (13 Apr, Shift 2)

Q41JEE Main 2023 · 13 APR · SHIFT 2Medium

Electric Charge & Coulomb's Law

Three point charges $q, -2q$ and $2q$ are placed on $x$ -axis at a distance $x = 0, x = \frac{3}{4}R$ and $x = R$ respectively from origin as shown. If $q = 2 \times 10^{-6}\text{ C}$ and $R = 2\text{ cm}$ , the magnitude of net force experienced by the charge $-2q$ is ___________ N.

Numerical answer type. Work it out, then open the answer.

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Answer: 5440

On $-2q$ (at $\tfrac{3}{4}R$ ): $q$ at distance $\tfrac{3}{4}R$ attracts it towards $-x$ : $\dfrac{k(q)(2q)}{(3R/4)^2} = \dfrac{32}{9}\dfrac{kq^2}{R^2}$ .
$2q$ at distance $\tfrac{R}{4}$ attracts it towards $+x$ : $\dfrac{k(2q)(2q)}{(R/4)^2} = 64\dfrac{kq^2}{R^2}$ .
Net $= \left(64 - \dfrac{32}{9}\right)\dfrac{kq^2}{R^2} = \dfrac{544}{9}\dfrac{kq^2}{R^2}$
$\dfrac{kq^2}{R^2} = \dfrac{9 \times 10^9 \times 4 \times 10^{-12}}{4 \times 10^{-4}} = 90\text{ N}$
Net force $= \dfrac{544}{9} \times 90 = 5440\text{ N}$ .

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