Question

Two point charges \(+8q\) and \(-2q\) are located at \(x=0\) and \(x=L\) respectively. What is the location of a point on the x-axis from the origin, at which the net electric field due to these two point charges is zero?

• A. \(8L\)
• B. \(4L\)
• C. \(2L\)
• D. \(\frac{L}{4}\)

Hint:

For unlike charges the field can vanish only outside the pair, on the side of the smaller charge.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Between unlike charges the fields add up, so a null point must lie outside, closer to the smaller charge \(-2q\), at some \(x>L\).

Step 2: Calculation
Setting the magnitudes equal: \(\frac{k\,8q}{x^{2}}=\frac{k\,2q}{(x-L)^{2}}\Rightarrow\frac{x}{x-L}=2\Rightarrow x=2L\).

Step 3: Compare the options
Check: at \(x=2L\), \(\frac{8q}{(2L)^{2}}=\frac{2q}{L^{2}}\), so the two fields are equal and opposite. The null point is at \(2L\).

Final Answer:
The net field is zero at \(x=2L\). \[ \boxed{\text{(C) }2L} \]