Question

A positive point charge is fixed at the origin. At some distance from it on the \(x\) axis, a point dipole is kept pointing in the \(+y\) direction. The force on the dipole is

• A. 0
• B. in the \(+y\) direction
• C. in the \(-y\) direction
• D. in the \(+x\) direction

Hint:

Use \(\vec F = (\vec p \cdot \nabla)\vec E\) for a dipole in an external field; since \(\vec p = p\hat y\), only \(\partial E_y/\partial y\) at the dipole's location matters, and for a point charge this comes out positive on the \(x\)-axis.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A point dipole with dipole moment \(\vec p\) sitting in a non-uniform external field \(\vec E\) feels a net force even though the total charge is zero, because the field is slightly different at its two ends. The force is given by \(\vec F = (\vec p \cdot \nabla)\vec E\), evaluated at the location of the dipole. Here the field is that of the point charge \(q\) fixed at the origin, and \(\vec p = p\hat y\) sits at a point \((d,0,0)\) on the \(x\)-axis.

Step 2: Key Formula or Approach:
1. Field of a point charge: \(\vec E(x,y,z) = \dfrac{kq}{r^3}(x,y,z)\), with \(r=\sqrt{x^2+y^2+z^2}\) and \(k=1/4\pi\epsilon_0\).
2. Since \(\vec p\) only has a \(y\)-component, the force reduces to \(\vec F = p\,\dfrac{\partial \vec E}{\partial y}\), evaluated at \((d,0,0)\).

Step 3: Detailed Explanation:
Differentiate each component of \(\vec E\) with respect to \(y\):
\[ \frac{\partial E_x}{\partial y} = kq\,\frac{\partial}{\partial y}\left(\frac{x}{r^3}\right) = -\frac{3kqxy}{r^5} \]
\[ \frac{\partial E_y}{\partial y} = kq\,\frac{\partial}{\partial y}\left(\frac{y}{r^3}\right) = kq\left(\frac{1}{r^3} - \frac{3y^2}{r^5}\right) \]
\[ \frac{\partial E_z}{\partial y} = kq\,\frac{\partial}{\partial y}\left(\frac{z}{r^3}\right) = -\frac{3kqzy}{r^5} \]
Now put in the dipole's location \((x,y,z)=(d,0,0)\), so \(r=d\) and \(y=z=0\). The first and third derivatives both carry a factor of \(y\) or \(z\), so they vanish at this point. Only the middle one survives:
\[ \left.\frac{\partial E_y}{\partial y}\right|_{(d,0,0)} = \frac{kq}{d^3} \]
So the gradient of \(\vec E\) at the dipole's location is nonzero only in the \(y\) direction, and:
\[ \vec F = p\left(0,\ \frac{kq}{d^3},\ 0\right) = \frac{kqp}{d^3}\hat y \]
Since \(q\), \(p\) and \(d^3\) are all positive, this force is a positive multiple of \(\hat y\).

Step 4: Why the other options are wrong.
The force is not zero (A), because the field of a point charge is genuinely non-uniform, so a dipole placed in it always feels some net pull unless its moment happens to be perpendicular to the local field gradient in just the right way, which is not the case here. It is not along \(-\hat y\) (C), because the sign of \(\partial E_y/\partial y\) at this point came out positive, not negative. It is not along \(\hat x\) (D), because both the \(x\) and \(z\) derivatives of \(\vec E\) vanish exactly at this symmetric point on the axis.

Step 5: Final Answer:
The dipole is pulled further along the \(y\)-direction it is already pointing in.\[ \boxed{\text{in the } +y \text{ direction}} \]