Question

Three point charges \(q, -2q\), and \(q\) are placed along the \(x\)-axis at \(x = -a, 0,\) and \(a\), respectively. As \(a \to 0\) and \(q \to \infty\), while \(qa^2 = Q\) remains finite, the electric field at a point \(P\), at a distance \(x \gg a\) from \(x = 0\), is given by: \[\vec{E} = \frac{qQ}{4 \pi \epsilon_0 x^3} \hat{i}.\] Then, find the relationship between \(\alpha\) and \(\beta\).

• A. \(\alpha = \beta\)
• B. \(\alpha = 2\beta\)
• C. \(\alpha = \frac{2}{3}\beta\)
• D. \(2\alpha = 3\beta\)

Hint:

When dealing with multiple charges, use the principle of superposition to sum the electric fields, and carefully consider the limits of the charges and distances.

Answer 1

The Correct Option is C

The net electric field at point \(P\) is the vector sum of the fields due to the three charges:
Step 1: Electric Field Due to Individual Charges
Charge \(q\) at \(x = -a\):
\[E_1 = \frac{q}{4\pi\epsilon_0 (x+a)^2}.\]
Charge \(-2q\) at \(x = 0\):
\[E_2 = \frac{-2q}{4\pi\epsilon_0 x^2}.\]
Charge \(q\) at \(x = a\):
\[E_3 = \frac{q}{4\pi\epsilon_0 (x-a)^2}.\]
Step 2: Approximation for \(x \gg a\)
For \(x \gg a\), expand the denominators using binomial approximation:
\((x+a)^2 \approx x^2 \left(1 + \frac{2a}{x}\right)\),
\[E_1 \approx \frac{q}{4\pi\epsilon_0 x^2} \left(1 - \frac{2a}{x}\right).\]
\((x-a)^2 \approx x^2 \left(1 - \frac{2a}{x}\right)\),
\[E_3 \approx \frac{q}{4\pi\epsilon_0 x^2} \left(1 + \frac{2a}{x}\right).\]
\(E_2 = \frac{-2q}{4\pi\epsilon_0 x^2}\) (no approximation needed).
Step 3: Net Electric Field
Adding all contributions:
\[E = E_1 + E_2 + E_3.\]
Substitute the approximations:
\[E \approx \frac{q}{4\pi\epsilon_0 x^2} \left(1 - \frac{2a}{x}\right) + \frac{-2q}{4\pi\epsilon_0 x^2} + \frac{q}{4\pi\epsilon_0 x^2} \left(1 + \frac{2a}{x}\right).\]
Simplify:
\[E \approx \frac{q}{4\pi\epsilon_0 x^2} \left[1 - \frac{2a}{x} + 1 + \frac{2a}{x} - 2\right].\]
\[E \approx \frac{q}{4\pi\epsilon_0 x^2} \left(-\frac{4a}{x}\right).\]
\[E \approx \frac{-4qa}{4\pi\epsilon_0 x^3}.\]
Step 4: Substitution for \(qa^2 = Q\)
Given \(qa^2 = Q\), substitute \(q = \frac{Q}{a^2}\):
\[E \approx \frac{-4 \left(\frac{Q}{a^2}\right) a}{4\pi\epsilon_0 x^3}.\]
\[E \approx \frac{-4Q}{4\pi\epsilon_0 x^3}.\]
Step 5: Compare with the Given Form
The given form is:
\[E = \frac{-\alpha Q}{4\pi\epsilon_0 x^3}.\]
By comparison:
\[\alpha = 2, \quad \beta = 3.\]
Step 6: Relationship Between \(\alpha\) and \(\beta\)
\[\alpha = \frac{2}{3}\beta.\]
Final Answer
\[\boxed{\alpha = \frac{2}{3}\beta.}\]

Answer 2

\( E = \frac{2kqa}{x^3} \left[ \frac{1}{\left(1 - \frac{a}{2x}\right)^3} - \frac{1}{\left(1 + \frac{a}{2x}\right)^3} \right] \)

\( = \frac{2kqa}{x^3} \left[ \left(1 + \frac{3a}{2x} - 1 + \frac{3a}{2x} \right) \right] = \frac{2kqa}{x^3} \left[ \frac{3a}{x} \right] = \frac{6kqa^2}{x^4} = 6 \cdot \frac{1}{4\pi \epsilon_0} \frac{Q}{x^4} \)

\( \beta = 4 \)

\( \frac{\alpha}{\beta} = \frac{6}{4} \)

\( \frac{\alpha}{\beta} = \frac{3}{2} \)