A point charge q1=4q0 is placed at origin. Another point charge q2=−q0 is placed at x = 12 cm. Charge of proton is q0 .The proton is placed on x axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is _____________cm.

Let proton is at point P, for net force zeroF1=F2⇒K4q0×q0(12+r)2=K×q0×q0r2⇒212+r=1r⇒r=12∴Position is 24cm from origin

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A point charge $q_{1} = 4 q_{0}$ is placed at origin. Another point charge $q_{2} = - q_{0}$ is placed at $x = 12 cm$ . Charge of proton is q₀ .The proton is placed on x axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is _____________cm.

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answer is 24.

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Detailed Solution

Let proton is at point P, for net force zero
$\begin{array}{l} F_{1} = F_{2} \\ \Rightarrow \frac{K 4 q_{0} \times q_{0}}{(12 + r)^{2}} = \frac{K \times q_{0} \times q_{0}}{r^{2}} \\ \Rightarrow \frac{2}{12 + r} = \frac{1}{r} \\ \Rightarrow r = 12 \end{array}$
$∴$ Position is 24cm from origin