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.)
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.)
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On the axis a dipole field goes as p/x^3; the fields cancel where the two magnitudes are equal.
Updated On: Oct 1, 2026
- \(2\) cm
- \(4\) cm
- \(6\) cm
- \(8\) cm
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The Correct Option is A
Solution and Explanation
Step 1: Field of a Dipole on Its Axis:
The magnitude is \(E=\dfrac{2kp}{x^3}\), proportional to \(\dfrac p{x^3}\). With the dipoles placed in opposite directions, the two fields point opposite between them, so they can cancel.
Step 2: First Case:
Let \(x_1\) be the distance from the dipole \(P\), so the distance from \(8P\) is \(24-x_1\).
\[ \frac P{x_1^3}=\frac{8P}{(24-x_1)^3}\Rightarrow\frac{24-x_1}{x_1}=2\Rightarrow x_1=8\ \text{cm} \]
Step 3: Second Case:
With \(27P\): \(\dfrac{24-x_2}{x_2}=\sqrt[3]{27}=3\Rightarrow x_2=6\) cm.
Step 4: Difference:
\(|x_2-x_1|=|6-8|=2\) cm. Option (A) is correct. Options 4, 6 and 8 cm would need \(x_2=4,\ 2,\ 0\) or similar values that do not satisfy the cube-root relation.
Final Answer:
The difference is 2 cm, option (A). \[ \boxed{\text{(A) } 2\ \text{cm}} \]
The magnitude is \(E=\dfrac{2kp}{x^3}\), proportional to \(\dfrac p{x^3}\). With the dipoles placed in opposite directions, the two fields point opposite between them, so they can cancel.
Step 2: First Case:
Let \(x_1\) be the distance from the dipole \(P\), so the distance from \(8P\) is \(24-x_1\).
\[ \frac P{x_1^3}=\frac{8P}{(24-x_1)^3}\Rightarrow\frac{24-x_1}{x_1}=2\Rightarrow x_1=8\ \text{cm} \]
Step 3: Second Case:
With \(27P\): \(\dfrac{24-x_2}{x_2}=\sqrt[3]{27}=3\Rightarrow x_2=6\) cm.
Step 4: Difference:
\(|x_2-x_1|=|6-8|=2\) cm. Option (A) is correct. Options 4, 6 and 8 cm would need \(x_2=4,\ 2,\ 0\) or similar values that do not satisfy the cube-root relation.
Final Answer:
The difference is 2 cm, option (A). \[ \boxed{\text{(A) } 2\ \text{cm}} \]
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