Two free point charges +q and +4q are a dista | vedclass

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Question

Let third charge $Q$ is located at a distance $x$ from $+q$ charge. For equilibrium force on it due to $+q$ charge and $+4 \mathrm{q}$ charge should b

Vedclass · Accepted Answer

$-\frac{4}{9}\,q$

Vedclass · Answer

Let third charge $Q$ is located at a distance $x$ from $+q$ charge. For equilibrium force on it due to $+q$ charge and $+4 \mathrm{q}$ charge should be equal in magnitude and directed oppositely.

$\frac{\mathrm{kqQ}}{\mathrm{x}^{2}}=\frac{\mathrm{kQ}(4 \mathrm{q})}{(\mathrm{R}-\mathrm{x})^{2}}$

or $\frac{\mathrm{R}-\mathrm{x}}{\mathrm{x}}=2$

or $\quad x=\frac{R}{3}$

for the entire system of free charges to be in equilibrium, the force on the $+q$ as well as $+4 q$ should also be zero. Consider force

$\frac{\mathrm{kq}(4 \mathrm{q})}{\mathrm{R}^{2}}+\frac{\mathrm{kQ}(4 \mathrm{q})}{(\mathrm{R}-\mathrm{x})^{2}}=0$

or $\frac{\mathrm{q}}{\mathrm{R}^{2}}+\frac{9 \mathrm{Q}}{4 \mathrm{R}^{2}}=0$

or $\quad \mathrm{Q}=-4 \mathrm{q} / 9$

The answer is $(4)$

Two free point charges $+q$ and $+4q$ are a distance $R$ apart. $A$ third charge is so placed that the entire system is in equilibrium. Then the third charge is :-

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