A point charge q1=4 q0 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.

Question

A point charge q1=4 q0 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. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/74cf898965d86709cac8b738b41c28d3-.png" /> (q0/x2)=(4 q0/(x+12)2) x+12=2 x x=12 Distance from origin =x+12=24 cm.

Q. A point charge q1​=4q0​ is placed at origin. Another point charge q2​=−q0​ is placed at x=12cm. 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.

x2q0​​=(x+12)24q0​​ x+12=2x x=12 Distance from origin =x+12=24cm.

$\frac{q _{0}}{x ^{2}} = \frac{4 q _{0}}{( x + 12 ) ^{2}}$
$x + 12 = 2 x$
$x = 12$
Distance from origin $= x + 12 = 24 c m$ .