Question

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:

• A. \( \frac{1 + \sqrt{3}}{r} \)
• B. \( \frac{r}{3(1 + \sqrt{3})} \)
• C. \( \frac{r}{1 + \sqrt{3}} \)
• D. \( r (1 + \sqrt{3}) \)

Answer 1

The Correct Option is C

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:

\(\frac{r}{1 + \sqrt{3}}\)

Answer 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:

\[ x = \frac{r}{\sqrt{3} + 1} \]