Two particles of charges $+e$ and $+2e$ are at 16 cm away from each other. Where should another charge q be placed between them, so that the system remains in equilibrium?
Show Hint
- 24 cm from $+e$
- 12.23 cm from $+e$
- 80 cm from $+e$
- 6.63 cm from $+e$
The Correct Option is D
Solution and Explanation
To find the position of a charge \( q \) such that the system remains in equilibrium, we apply the principle that the net force acting on the charge must be zero. Here, one charge is \( +e \) and the other is \( +2e \), separated by a distance of 16 cm.
Let the distance of charge \( q \) from the charge \( +e \) be \( x \), and hence the distance from \( +2e \) will be \( 16 - x \) cm.
Using Coulomb's Law, the force on charge \( q \) due to \( +e \) is given by:
\(F_1 = \frac{k \cdot e \cdot q}{x^2}\)
And the force on charge \( q \) due to \( +2e \) is given by:
\(F_2 = \frac{k \cdot 2e \cdot q}{(16 - x)^2}\)
For equilibrium, these forces must be equal in magnitude:
\(\frac{k \cdot e \cdot q}{x^2} = \frac{k \cdot 2e \cdot q}{(16 - x)^2}\)
Cancelling common terms:
\(\frac{1}{x^2} = \frac{2}{(16 - x)^2}\)
Cross-multiplying gives:
\((16 - x)^2 = 2x^2\)
Expanding and solving:
\(256 - 32x + x^2 = 2x^2\)
Rearranging gives:
\(x^2 + 32x - 256 = 0\)
Now using the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \( a = 1 \), \( b = 32 \), and \( c = -256 \)
\(x = \frac{-32 \pm \sqrt{32^2 - 4 \times 1 \times (-256)}}{2 \times 1}\)
\(x = \frac{-32 \pm \sqrt{1024 + 1024}}{2}\)
\(x = \frac{-32 \pm \sqrt{2048}}{2}\)
\(x = \frac{-32 \pm 45.25}{2}\)
Considering the positive root as distance cannot be negative:
\(x \approx \frac{-32 + 45.25}{2} \approx 6.63 \, \text{cm}\)
Thus, the charge \( q \) should be placed approximately 6.63 cm from \( +e \) for the system to be in equilibrium.
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