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: Model the dipole as two nearby point charges.
Instead of using the dipole-force formula directly, replace the point dipole by two real charges: $+q_d$ at $(d, \delta/2, 0)$ and $-q_d$ at $(d, -\delta/2, 0)$, with $\delta$ very small, so that the dipole moment is $p = q_d \delta$ pointing along $+y$ (from the negative charge to the positive one), matching the direction given in the problem.

Step 2: Write the force from the fixed charge $Q$ on each of the two charges.
Let $Q$ (positive) sit at the origin. The distance from $Q$ to each of the two charges is nearly the same for small $\delta$:
\[ r_+ = \sqrt{d^2 + \delta^2/4} \approx d, \qquad r_- = \sqrt{d^2+\delta^2/4}\approx d \]
The Coulomb force on $+q_d$ is $\vec F_+ = \dfrac{kQq_d}{r_+^3}(d,\ \delta/2,\ 0)$, and on $-q_d$ it is $\vec F_- = -\dfrac{kQq_d}{r_-^3}(d,\ -\delta/2,\ 0)$.

Step 3: Add the two forces and keep only the leading order in $\delta$.
Since $r_+^3$ and $r_-^3$ agree with $d^3$ up to corrections of order $\delta^2$, we can replace both by $d^3$ when adding the forces to leading order in $\delta$:
\[ \vec F_+ + \vec F_- \approx \frac{kQq_d}{d^3}\Big[(d,\ \delta/2,\ 0) - (d,\ -\delta/2,\ 0)\Big] = \frac{kQq_d}{d^3}(0,\ \delta,\ 0) \]
Notice the $x$-components of the two forces are equal and cancel out when we form $\vec F_+ + \vec F_-$ (both contribute $kQq_d d/d^3$, but with opposite overall sign from the charges), while the $y$-components add constructively.

Step 4: Write the total force in terms of $p$.
\[ \vec F = \frac{kQq_d\delta}{d^3}\hat y = \frac{kQp}{d^3}\hat y \]
since $p=q_d\delta$. This is the same result as the gradient formula, but reached by directly adding up two real Coulomb forces instead of differentiating the field.

Step 5: Interpret the sign.
The $+q_d$ end sits slightly further along $+y$ where the field of $Q$ has a slightly larger $y$-component than at the $-q_d$ end (because moving along $+y$ at fixed $x=d$ increases the radial distance and rotates the field vector to pick up more $y$-component up to this order), so the pull on $+q_d$ in $+y$ outweighs the pull on $-q_d$ in $-y$, giving a net push of the whole dipole toward $+y$.

Final Answer:
Adding the two Coulomb forces gives a net force along $+y$, the same direction the dipole already points in.\[ \boxed{\text{in the } +y \text{ direction}} \]