Question:
Three charges \(+5q\), \(Q\) and \(-2q\) are kept along a straight line in the same order such that, \(+5q\) and \(-2q\) charges are at a distance of \(\frac{2r}{3}\) and \(\frac{r}{3}\) from the charge \(Q\) respectively. If the net force on the charge \(-2q\) is zero, then \(Q\) is
Three charges \(+5q\), \(Q\) and \(-2q\) are kept along a straight line in the same order such that, \(+5q\) and \(-2q\) charges are at a distance of \(\frac{2r}{3}\) and \(\frac{r}{3}\) from the charge \(Q\) respectively. If the net force on the charge \(-2q\) is zero, then \(Q\) is
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For electrostatic equilibrium of a charge, first decide the directions of forces and then equate their magnitudes. The sign of the unknown charge is decided by whether the force must be attractive or repulsive.
Updated On: Jun 26, 2026
- \(+\dfrac{5}{9}q\)
- \(-\dfrac{5}{9}q\)
- \(3q\)
- \(-3q\)
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The Correct Option is B
Solution and Explanation
Step 1: Understand the arrangement of charges.
The charges are placed in the order \[ +5q,\quad Q,\quad -2q \] The distance between \(+5q\) and \(Q\) is \[ \frac{2r}{3} \] The distance between \(Q\) and \(-2q\) is \[ \frac{r}{3} \] Therefore, the distance between \(+5q\) and \(-2q\) is \[ \frac{2r}{3}+\frac{r}{3}=r \]
Step 2: Force on \(-2q\) due to \(+5q\).
Since \(+5q\) and \(-2q\) are opposite charges, the force is attractive.
So, the force on \(-2q\) due to \(+5q\) is towards \(+5q\), that is, towards the left.
Magnitude of this force is \[ F_1=\frac{k(5q)(2q)}{r^2} \] \[ F_1=\frac{10kq^2}{r^2} \]
Step 3: Direction of force due to \(Q\).
For the net force on \(-2q\) to be zero, the force due to \(Q\) must be opposite to \(F_1\).
So, force due to \(Q\) on \(-2q\) must be towards the right.
Since \(Q\) is to the left of \(-2q\), this is possible only if \(Q\) is negative, so that \(Q\) and \(-2q\) repel each other.
Step 4: Force on \(-2q\) due to \(Q\).
Distance between \(Q\) and \(-2q\) is \[ \frac{r}{3} \] Magnitude of force due to \(Q\) is \[ F_2=\frac{k|Q|(2q)}{\left(\frac{r}{3}\right)^2} \] \[ F_2=\frac{2k|Q|q}{\frac{r^2}{9}} \] \[ F_2=\frac{18k|Q|q}{r^2} \]
Step 5: Apply condition of zero net force.
For net force on \(-2q\) to be zero, \[ F_1=F_2 \] Thus, \[ \frac{10kq^2}{r^2} = \frac{18k|Q|q}{r^2} \] Cancelling common terms, \[ 10q=18|Q| \] \[ |Q|=\frac{10q}{18} \] \[ |Q|=\frac{5q}{9} \] Since \(Q\) must be negative, \[ Q=-\frac{5q}{9} \]
Step 6: Final conclusion.
Therefore, \[ \boxed{Q=-\frac{5q}{9}} \]
The charges are placed in the order \[ +5q,\quad Q,\quad -2q \] The distance between \(+5q\) and \(Q\) is \[ \frac{2r}{3} \] The distance between \(Q\) and \(-2q\) is \[ \frac{r}{3} \] Therefore, the distance between \(+5q\) and \(-2q\) is \[ \frac{2r}{3}+\frac{r}{3}=r \]
Step 2: Force on \(-2q\) due to \(+5q\).
Since \(+5q\) and \(-2q\) are opposite charges, the force is attractive.
So, the force on \(-2q\) due to \(+5q\) is towards \(+5q\), that is, towards the left.
Magnitude of this force is \[ F_1=\frac{k(5q)(2q)}{r^2} \] \[ F_1=\frac{10kq^2}{r^2} \]
Step 3: Direction of force due to \(Q\).
For the net force on \(-2q\) to be zero, the force due to \(Q\) must be opposite to \(F_1\).
So, force due to \(Q\) on \(-2q\) must be towards the right.
Since \(Q\) is to the left of \(-2q\), this is possible only if \(Q\) is negative, so that \(Q\) and \(-2q\) repel each other.
Step 4: Force on \(-2q\) due to \(Q\).
Distance between \(Q\) and \(-2q\) is \[ \frac{r}{3} \] Magnitude of force due to \(Q\) is \[ F_2=\frac{k|Q|(2q)}{\left(\frac{r}{3}\right)^2} \] \[ F_2=\frac{2k|Q|q}{\frac{r^2}{9}} \] \[ F_2=\frac{18k|Q|q}{r^2} \]
Step 5: Apply condition of zero net force.
For net force on \(-2q\) to be zero, \[ F_1=F_2 \] Thus, \[ \frac{10kq^2}{r^2} = \frac{18k|Q|q}{r^2} \] Cancelling common terms, \[ 10q=18|Q| \] \[ |Q|=\frac{10q}{18} \] \[ |Q|=\frac{5q}{9} \] Since \(Q\) must be negative, \[ Q=-\frac{5q}{9} \]
Step 6: Final conclusion.
Therefore, \[ \boxed{Q=-\frac{5q}{9}} \]
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