Question

Two electric dipoles of moment \(P\) and \(8P\) are placed in opposite directions on a line at a distance of \(24\) cm. The electric field will be zero at the point between the dipoles whose distance from the dipole of moment \(P\) will be \(x_1\). When \(8P\) is replaced by \(27P\) keeping all other quantities the same, the distance \(x_1\) now becomes \(x_2\). The difference \(|x_2-x_1|\) is ( All the distances are measured from the centers of the dipole.)

• A. \(2\) cm
• B. \(4\) cm
• C. \(6\) cm
• D. \(8\) cm

Hint:

On the axis a dipole field goes as p/x^3; the fields cancel where the two magnitudes are equal.

Answer 1

The Correct Option is A

Step 1: Equate Magnitudes:
The cancellation condition is $\dfrac{x}{24-x}=\left(\dfrac{P}{8P}\right)^{1/3}=\dfrac12$, taking the distance $x$ from the small dipole.

Step 2: Solve Both Cases:
Case 1: $\dfrac x{24-x}=\dfrac12\Rightarrow2x=24-x\Rightarrow x_1=8$. Case 2: $\dfrac x{24-x}=\left(\dfrac1{27}\right)^{1/3}=\dfrac13\Rightarrow3x=24-x\Rightarrow x_2=6$.

Step 3: Answer:
$|x_2-x_1|=2$ cm. Option (A).

Final Answer:
Option (A). \[ \boxed{\text{(A) } 2\ \text{cm}} \]