Electrostatics Question 9
Question 9 - 29 January - Shift 1
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 $q_0$. The proton is placed on $x$-axis so that the electrostatic force on the proton in zero. In this situation, the position of the proton from the origin is $cm$.
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Answer: (24)
Solution:
Formula: Coulomb Force
$\frac{q_0}{x^{2}}=\frac{4 q_0}{(x+12)^{2}}$
$x+12=2 x$
$x=12$
