Question

Then, Net force on the dipole : $F_{net} = qE(x + dx) - qE(x)$

$F_{net} = q \frac{E(x+dx) - E(x)}{dx} (dx)$; here $(q(dx) = P)$ $\therefore F_{net} = P \left( \frac{dE}{dx} \right)$

Solved Examples

xample 91. A short dipole is placed on the axis of a uniformly charged ring (total c distance $\frac{R}{\sqrt{2}}$ from centre of ring as shown in figure. Find the Force

$\therefore F = P \left( \frac{dE}{dx} \right)$

$\therefore F = P \frac{d}{dx} \left( \frac{KQx}{(R^2 + x^2)^{3/2}} \right) ; (at x = \frac{R}{\sqrt{2}})$

Solving we get, F = 0 ??

Answer 1

0

Answer 2

The problem asks to determine the net force on a short electric dipole placed on the axis of a uniformly charged ring. The dipole is located at a distance $x = \frac{R}{\sqrt{2}}$ from the center of the ring, where $R$ is the radius of the ring.

The net force on a short dipole with dipole moment $P$ in a non-uniform electric field $E(x)$ along the x-axis is given by: $F_{net} = P \left( \frac{dE}{dx} \right)$

The electric field $E(x)$ due to a uniformly charged ring of total charge $Q$ and radius $R$ at a distance $x$ along its axis is: $E(x) = \frac{KQx}{(R^2 + x^2)^{3/2}}$ where $K = \frac{1}{4\pi\epsilon_0}$ is Coulomb's constant.

To find the force, we need to calculate the derivative of $E(x)$ with respect to $x$: $\frac{dE}{dx} = \frac{d}{dx} \left( \frac{KQx}{(R^2 + x^2)^{3/2}} \right)$ We can use the quotient rule, or rewrite $E(x)$ as $KQx(R^2 + x^2)^{-3/2}$ and use the product rule. Let's use the product rule for $u=x$ and $v=(R^2+x^2)^{-3/2}$: $\frac{dE}{dx} = KQ \left[ \frac{d}{dx}(x) \cdot (R^2 + x^2)^{-3/2} + x \cdot \frac{d}{dx}((R^2 + x^2)^{-3/2}) \right]$ $\frac{dE}{dx} = KQ \left[ 1 \cdot (R^2 + x^2)^{-3/2} + x \cdot \left( -\frac{3}{2}(R^2 + x^2)^{-5/2} \cdot 2x \right) \right]$ $\frac{dE}{dx} = KQ \left[ (R^2 + x^2)^{-3/2} - 3x^2(R^2 + x^2)^{-5/2} \right]$ Factor out $(R^2 + x^2)^{-5/2}$: $\frac{dE}{dx} = KQ (R^2 + x^2)^{-5/2} \left[ (R^2 + x^2) - 3x^2 \right]$ $\frac{dE}{dx} = \frac{KQ(R^2 - 2x^2)}{(R^2 + x^2)^{5/2}}$

Now, we need to evaluate $\frac{dE}{dx}$ at the given position $x = \frac{R}{\sqrt{2}}$. Substitute $x = \frac{R}{\sqrt{2}}$ into the expression for $\frac{dE}{dx}$: $R^2 - 2x^2 = R^2 - 2 \left( \frac{R}{\sqrt{2}} \right)^2$ $R^2 - 2 \left( \frac{R^2}{2} \right) = R^2 - R^2 = 0$

Since the numerator $(R^2 - 2x^2)$ becomes zero at $x = \frac{R}{\sqrt{2}}$, the entire derivative $\frac{dE}{dx}$ at this point is zero: $\frac{dE}{dx} \Big|_{x = R/\sqrt{2}} = \frac{KQ(0)}{(R^2 + (R/\sqrt{2})^2)^{5/2}} = 0$

Finally, the net force on the dipole is: $F_{net} = P \left( \frac{dE}{dx} \right) = P \times 0 = 0$

This confirms that the net force on the dipole at $x = \frac{R}{\sqrt{2}}$ is zero. This point is significant because it is where the electric field due to the ring on its axis reaches its maximum magnitude, and thus its derivative is zero.

The final answer is $\boxed{0}$.