Animated Solution for Physics - Electrostatics: Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q$. The distance of the point P from the centre of the rod is $a=\frac{\sqrt{3}}{2}L$.

The beauty of symmetry in electrostatics often turns seemingly complex integration problems into elegant geometric puzzles. In this problem, we are tasked with finding the electric field at a specific point on the perpendicular bisector of a uniformly charged finite wire. Let's embark on this journey and see how geometry simplifies the physics.

Imagine a uniformly charged thin wire of length $L$ carrying a total charge $Q$. We are interested in a point $P$ located on its perpendicular bisector at a distance $a=\frac{\sqrt{3}}{2}L$ from the center of the wire.

Because point $P$ lies exactly on the perpendicular bisector, the setup is highly symmetric. The vertical components of the electric field produced by the upper half of the wire perfectly cancel out the vertical components produced by the lower half. Thus, the net electric field $\vec{E}$ points strictly horizontally, away from the wire (assuming $Q$ is positive).

For a finite charged wire, the net electric field at a point on its perpendicular bisector is given by the standard derived formula:

Here, $k=\frac{1}{4\pi {\epsilon }_0}$ is Coulomb's constant, $\lambda =\frac{Q}{L}$ is the linear charge density, $a$ is the perpendicular distance to the point, and ${\varphi }_1$ and ${\varphi }_2$ are the angles subtended by the two ends of the wire at point $P$.

To use our master equation, we need to find the values of $\sin {\varphi }_1$ and $\sin {\varphi }_2$. Let's look at the right-angled triangle formed by the center of the wire, one of its ends, and point $P$.

The base of this triangle is $a=\frac{\sqrt{3}}{2}L$, and the height is half the length of the wire, which is $\frac{L}{2}$. To find the sine of the angle $\varphi$, we first need the hypotenuse $r$. Using the Pythagorean theorem:

Substituting the given value of $a$:

It turns out the hypotenuse is exactly equal to the length of the wire $L$! This makes calculating the sine of the angles incredibly simple. Since the point is on the bisector, ${\varphi }_1={\varphi }_2=\varphi$.

Now, we have all the pieces of the puzzle. Let's substitute everything back into our master equation:

The sum of the sine terms beautifully collapses to $1$. The factor of $2$ in the denominator of $a$ flips up to the numerator:

Simplifying the constants, we arrive at our final, elegant result:

By leveraging the standard formula and basic trigonometry, we bypassed a tedious integration. As a thought experiment, consider what happens if the wire becomes infinitely long ($L\to \infty$). The angles ${\varphi }_1$ and ${\varphi }_2$ would approach $90^{\circ }$, making their sines equal to $1$. The formula would then reduce to $E=\frac{\lambda }{2\pi {\epsilon }_{0}a}$, which perfectly matches the result obtained using Gauss's Law for an infinite wire!