Two fixed point charges $+ 4 e$ and $+ e$ are separated by a distance a. Where should the third point charge be placed from $+ 4 e$ charge for it to be in equilibrium?

\frac{2}{3} a

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\frac{4}{3} a

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\frac{1}{3} a

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None of these

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Solution

The correct option is A $\frac{2}{3} a$

The third charge q is in equilibrium only when
force on it due to $+ 4 e =$ force due to $+ e$

i.e $\frac{k (4 e) q}{x^{2}} = \frac{k (e) q}{(a - x)^{2}}$

or $4 (a - x)^{2} = x^{2}$

or $2 (a - x) = x$

or $2 a - 2 x = x \Rightarrow x = \frac{2 a}{3}$

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Similar questions

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Two fixed point charges +4e and +e are separated by a distance a. Where should the third point charge be placed from +4e charge for it to be in equilibrium?

The correct option is A 23a The third charge q is in equilibrium only when force on it due to +4e= force due to +e i.e k(4e)qx2=k(e)q(a−x)2 or 4(a−x)2=x2 or 2(a−x)=x or 2a−2x=x⇒x=2a3

Two fixed point charges $+ 4 e$ and $+ e$ are separated by a distance a. Where should the third point charge be placed from $+ 4 e$ charge for it to be in equilibrium?

The correct option is A $\frac{2}{3} a$

The third charge q is in equilibrium only when
force on it due to $+ 4 e =$ force due to $+ e$

i.e $\frac{k (4 e) q}{x^{2}} = \frac{k (e) q}{(a - x)^{2}}$

or $4 (a - x)^{2} = x^{2}$

or $2 (a - x) = x$

or $2 a - 2 x = x \Rightarrow x = \frac{2 a}{3}$