Three charges +3q,Q,and +q are placed in a st | vedclass

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(B) Let the charges be $q_1 = +3q$ at $x = 0$,$q_2 = Q$ at $x = \frac{\ell}{2}$,and $q_3 = +q$ at $x = \ell$.For the net force on the charge $+q$ at $

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$-\frac{3}{4}$

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(B) Let the charges be $q_1 = +3q$ at $x = 0$,$q_2 = Q$ at $x = \frac{\ell}{2}$,and $q_3 = +q$ at $x = \ell$.For the net force on the charge $+q$ at $x = \ell$ to be zero,the sum of the electrostatic forces exerted by $q_1$ and $q_2$ must be zero.Using Coulomb's Law,$F = \frac{k q_1 q_2}{r^2}$.The force exerted by $q_1$ on $q_3$ is $F_1 = \frac{k(3q)(q)}{\ell^2}$.The force exerted by $q_2$ on $q_3$ is $F_2 = \frac{k(Q)(q)}{(\ell/2)^2} = \frac{kQq}{\ell^2/4} = \frac{4kQq}{\ell^2}$.For the net force to be zero,$F_1 + F_2 = 0$,so $\frac{3kq^2}{\ell^2} + \frac{4kQq}{\ell^2} = 0$.Dividing by $\frac{kq}{\ell^2}$,we get $3q + 4Q = 0$.Therefore,$4Q = -3q$,which gives $Q = -\frac{3}{4}q$.Comparing this with $Q = xq$,we find $x = -\frac{3}{4}$.

Three charges $+3q$,$Q$,and $+q$ are placed in a straight line of length $\ell$ at points at distances $0$,$\frac{\ell}{2}$,and $\ell$ respectively. The value of $Q$ in order to have the net force on $+q$ to be zero is $Q = xq$. The value of $x$ is:

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