Question

The distance between charges \(+q\) and \(-q\) is \(2l\) and between \(+2q\) and \(-2q\) is \(4l\). The electrostatic potential at point \(P\) at a distance \(r\) from center \(O\) is \(-\alpha \left[\frac{q}{r^2}\right] \times 10^9 \, \text{V}\), where the value of \(\alpha\) is ______.
(Use \(\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2}\))

Answer 1

Correct Answer: 27

Provided Information:

The potential due to a point charge is defined as: \[ V_q = \frac{kP \cos \theta}{r^2} \], where \( k \) represents the Coulomb constant, \( P \) is the charge, and \( r \) denotes the distance.

Step 1: Potential due to charge \( q \)

The potential generated by charge \( q \) is calculated using: \[ V_q = \frac{K q l}{r^2} \quad \cdots \, (i) \]

Step 2: Potential due to charge \( 2q \)

The potential arising from charge \( 2q \) is determined as: \[ V_{2q} = K[2q(4)] \cos 120^\circ = 8 K q l \cdot \left( -\frac{1}{2} \right) \] which simplifies to \[ V_{2q} = \frac{-4 K q l}{r^2} \quad \cdots \, (ii) \]

Step 3: Combining potentials (i) and (ii)

The total potential, \( V_{\text{net}} \), is the sum of the potentials from charges \( q \) and \( 2q \): \[ V_{\text{net}} = \frac{K q l}{r^2} - \frac{4 K q l}{r^2} = \frac{-3 K q l}{r^2} \]

Step 4: Simplification and numerical evaluation

\[ V_{\text{net}} = -3 \left[ \frac{q l}{r^2} \right] \times 9 \times 10^9 \] which yields \[ = -27 \left[ \frac{q l}{r^2} \right] \times 10^9 \]

Step 5: Determining \( \alpha \)

Given that \( \alpha = 27 \), it is established that: \[ \alpha = 27 \]

Conclusion:

\[ \boxed{\alpha = 27} \]