Charges 4Q,q,and Q are placed along the x-axi | vedclass

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Question

(D) The force on charge $Q$ at $x = l$ is due to the charges $4Q$ at $x = 0$ and $q$ at $x = l/2$.Using Coulomb's Law,the force exerted by $4Q$ on $Q$

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$-Q$

Vedclass · Answer

(D) The force on charge $Q$ at $x = l$ is due to the charges $4Q$ at $x = 0$ and $q$ at $x = l/2$.Using Coulomb's Law,the force exerted by $4Q$ on $Q$ is $F_1 = \frac{1}{4\pi\varepsilon_0} \frac{(4Q)(Q)}{l^2}$.The force exerted by $q$ on $Q$ is $F_2 = \frac{1}{4\pi\varepsilon_0} \frac{qQ}{(l/2)^2}$.For the net force on $Q$ to be zero,$F_1 + F_2 = 0$.$\frac{1}{4\pi\varepsilon_0} \frac{4Q^2}{l^2} + \frac{1}{4\pi\varepsilon_0} \frac{qQ}{l^2/4} = 0$.$\frac{4Q^2}{l^2} + \frac{4qQ}{l^2} = 0$.Dividing by $\frac{4Q}{l^2}$ (assuming $Q 
eq 0$ and $l 
eq 0$),we get $Q + q = 0$.Therefore,$q = -Q$.

Charges $4Q$,$q$,and $Q$ are placed along the $x$-axis at positions $x = 0$,$x = l/2$,and $x = l$,respectively. Find the value of $q$ such that the net force on charge $Q$ is zero.

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