Three charges 4q,,Q and q are in a straight line in position of 0 , l/2 and l respectively. resultan

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1. Electric Charges and Fields

normal

Three charges $4q,\,Q$ and $q$ are in a straight line in the position of $0$, $l/2$ and $l$ respectively. The resultant force on $q$ will be zero, if $Q = $

$-q$

$ - \,2q$

$ - \frac{q}{2}$

$4q$

Solution

(a) The force between $4q$ and $q$; ${F_1} = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{4q \times q}}{{{l^2}}}$
The force between $Q$ and $q$; ${F_2} = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Q \times q}}{{{{(l/2)}^2}}}$
We want ${F_1} + {F_2} = 0$ or $\frac{{4{q^2}}}{{{l^2}}} = – \frac{{4Qq}}{{{l^2}}}$ $==>$ $Q = – q$

Standard 12

Physics

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(KVPY-2016)

Three charges $4q,\,Q$ and $q$ are in a straight line in the position of $0$, $l/2$ and $l$ respectively. The resultant force on $q$ will be zero, if $Q = $

Solution

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