Question

Two point charges \(8q\) and \(2q\) are located at \(x=0\) and \(x=L\) respectively. The location of a point on the X-axis from origin, at which the net electric field is zero, is \(x\). Value of \(x\) is equal to

• A. \(\frac{2L}{3}\)
• B. \(\frac{3L}{2}\)
• C. \(\frac{L}{3}\)
• D. \(\frac{L}{2}\)

Hint:

The fields of the two like charges cancel between them, nearer the smaller charge.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Both charges are of the same sign, so their fields cancel at a point between them.

Step 2: Equation
At distance \(x\) from \(8q\): \(\dfrac{k\,8q}{x^2}=\dfrac{k\,2q}{(L-x)^2}\), so \(\dfrac{4}{x^2}=\dfrac{1}{(L-x)^2}\).

Step 3: Calculation
Taking the square root, \(\dfrac{2}{x}=\dfrac{1}{L-x}\), so \(2L-2x=x\).

Step 4: Result
\(x=\dfrac{2L}{3}\), which lies between 0 and \(L\).

Final Answer:
The field is zero at x equal to 2L/3 from the origin. \[ \boxed{\text{(A) }\dfrac{2L}{3}} \]