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

Given:

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

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

The potential due to the charge \( q \) is given by: \[ V_q = \frac{K q l}{r^2} \quad \cdots \, (i) \]

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

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

Step 3: Using (i) and (ii)

The net potential \( V_{\text{net}} \) is obtained by adding the potentials due to \( q \) 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 calculation

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

Step 5: Finding \( \alpha \)

Given \( \alpha = 27 \), we conclude: \[ \alpha = 27 \]

Final Answer:

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

Answer 2

Solution: The problem involves finding the net dipole moment and the resulting potential at a point due to two pairs of charges arranged as specified.

The charges are arranged in two pairs: - Pair 1: \(+q\) and \(-q\) separated by a distance of \(2l\). - Pair 2: \(+2q\) and \(-2q\) separated by a distance of \(4l\).

The dipole moment \(P\) for a pair of charges \(+Q\) and \(-Q\) separated by distance \(d\) is given by:

\[ P = Q \cdot d. \]

For Pair 1:

\[ P_1 = q \cdot (2l) = 2ql. \]

For Pair 2:

\[ P_2 = 2q \cdot (4l) = 8ql. \]

The two dipole moments \(P_1\) and \(P_2\) are positioned at an angle of \(120^\circ\) relative to each other. The magnitude of the resultant dipole moment \(P_{\text{net}}\) can be found using the vector addition formula:

\[ P_{\text{net}} = \sqrt{P_1^2 + P_2^2 + 2P_1P_2\cos\theta}. \]

Substituting \(P_1 = 2ql\), \(P_2 = 8ql\), and \(\theta = 120^\circ\):

\[ P_{\text{net}} = \sqrt{(2ql)^2 + (8ql)^2 + 2 \cdot (2ql) \cdot (8ql) \cdot \cos 120^\circ}. \]

Since \(\cos 120^\circ = -\frac{1}{2}\):

\[ P_{\text{net}} = \sqrt{4q^2l^2 + 64q^2l^2 + 2 \cdot (2ql) \cdot (8ql) \cdot \left(-\frac{1}{2}\right)}. \] \[ P_{\text{net}} = \sqrt{4q^2l^2 + 64q^2l^2 - 16q^2l^2}. \] \[ P_{\text{net}} = \sqrt{52q^2l^2} = \sqrt{36q^2l^2} = 6ql. \]

The potential \(V\) at a point on the axis of a dipole at a distance \(r\) from the center is given by:

\[ V = \frac{KP \cos \theta}{r^2}. \]

Here, \(K = \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2\text{C}^{-2}\) and \(\theta = 120^\circ\).

Substitute \(K\), \(P_{\text{net}} = 6ql\), and \(\cos 120^\circ = -\frac{1}{2}\):

\[ V = \frac{9 \times 10^9 \cdot 6ql \cdot \left(-\frac{1}{2}\right)}{r^2}. \] \[ V = \frac{-27 \times 10^9 \cdot ql}{r^2}. \]

Thus, the value of \(\alpha\) in the potential expression is:

\[ \alpha = 27. \]