Two charges \( q \) and \( 3q \) are separated by a distance \( r \) in air. At a distance \( x \) from charge \( q \), the resultant electric field is zero. The value of \( x \) is:
- \( \frac{1 + \sqrt{3}}{r} \)
- \( \frac{r}{3(1 + \sqrt{3})} \)
- \( \frac{r}{1 + \sqrt{3}} \)
- \( r (1 + \sqrt{3}) \)
The Correct Option is C
Approach Solution - 1
To solve this problem, we need to find the position \( x \) from the charge \( q \) at which the resultant electric field becomes zero. We have two point charges: \( q \) and \( 3q \), separated by a distance \( r \).
The formula for the electric field \( E \) due to a point charge is given by:
\(E = \frac{k \cdot Q}{d^2}\)
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( d \) is the distance from the charge.
At a distance \( x \) from charge \( q \), the electric field due to charge \( q \) is:
\(E_1 = \frac{k \cdot q}{x^2}\)
The electric field due to charge \( 3q \) located at a distance \( r-x \) (since the total separation between the charges is \( r \)) is:
\(E_2 = \frac{k \cdot 3q}{(r-x)^2}\)
For the net electric field to be zero at point \( x \), the magnitudes of these fields must be equal:
\(\frac{k \cdot q}{x^2} = \frac{k \cdot 3q}{(r-x)^2}\)
After equating and simplifying, we have:
\(\frac{1}{x^2} = \frac{3}{(r-x)^2}\)
Take the square root on both sides:
\(\frac{1}{x} = \frac{\sqrt{3}}{r-x}\)
Cross-multiply to solve for \( x \):
\(r\sqrt{3} = x\sqrt{3} + x\)
\(x(\sqrt{3} + 1) = r\sqrt{3}\)
From this, solve for \( x \):
\(x = \frac{r\sqrt{3}}{\sqrt{3} + 1}\)
Multiplying the numerator and the denominator by the conjugate of the denominator (\(\sqrt{3} - 1\)), we have:
\(x = \frac{r\sqrt{3}(\sqrt{3} - 1)}{3 - 1}\)
\(x = \frac{r(3 - \sqrt{3})}{2}\)
Rewriting using simplification leads to:
\(x = \frac{r}{1 + \sqrt{3}}\)
Thus, the correct answer is:
\(\frac{r}{1 + \sqrt{3}}\)
Approach Solution -2
Given two charges \( q \) and \( 3q \) separated by a distance \( r \). The electric field at a point \( x \) from charge \( q \) where the net electric field is zero is:
\[ \vec{E}_{\text{net}} = 0 \]
Equating the electric fields due to both charges:
\[ k \frac{q}{x^2} = k \frac{3q}{(r - x)^2} \]
Simplifying:
\[ (r - x)^2 = 3x^2 \] \[ r - x = \sqrt{3}x \]
Rearranging gives:
\[ x = \frac{r}{\sqrt{3} + 1} \]
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