(A) Let the proton be placed at a distance $r$ from the origin. The force due to $q_1$ at the origin is $F_1 = \frac{k (4q_0) q_0}{r^2}$ (repulsive,to

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(A) Let the proton be placed at a distance $r$ from the origin. The force due to $q_1$ at the origin is $F_1 = \frac{k (4q_0) q_0}{r^2}$ (repulsive,towards the right).The force due to $q_2$ at $x = 12 \, cm$ is $F_2 = \frac{k q_0 q_0}{(r - 12)^2}$ (attractive,towards the left).For the net force to be zero,$F_1 = F_2$:$\frac{4 k q_0^2}{r^2} = \frac{k q_0^2}{(r - 12)^2}$Taking the square root on both sides:$\frac{2}{r} = \frac{1}{r - 12}$$2(r - 12) = r$$2r - 24 = r$$r = 24 \, cm$.Thus,the proton is at a distance of $24 \, cm$ from the origin.

Class 12 Physics Electric Charges and Fields Electrostatic Force and Coulombs Law

Medium

$A$ point charge $q_1 = 4 q_0$ is placed at the origin. Another point charge $q_2 = -q_0$ is placed at $x = 12 \, cm$. The charge of a proton is $q_0$. $A$ proton is placed on the $x$-axis such that the net electrostatic force on it is zero. In this situation,the position of the proton from the origin is $.......... \, cm$.

JEE MAIN 2023 All JEE MAIN

A
$24$
B
$23$
C
$22$
D
$20$

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Electric Charges and Fields

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Electrostatic Force and Coulombs Law

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Two spheres carrying charges $+6 \mu C$ and $+9 \mu C$ are separated by a distance $d$ and experience a force of repulsion $F$. When a charge of $-3 \mu C$ is added to both spheres and they are kept at the same distance as before,the new force of repulsion is:

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