An infinitely long uniform line charge distribution of charge per unit length lambda lies parallel t

Q: Why don't we integrate the electric field over the area?

Integration works perfectly fine, but it's mathematically tedious and prone to calculation errors. Symmetry provides a much more elegant, intuitive, and faster solution, which is exactly what JEE tests.

Q: Why is the flux proportional to the 2D planar angle and not the 3D solid angle?

Because the line charge is infinitely long, the electric field is purely radial in cylindrical coordinates. The problem has translational symmetry along the y-axis, effectively reducing it to a 2D problem where the flux fraction is determined solely by the planar angle.

Q: Does the distance of the rectangle from the origin matter?

Yes, absolutely. The specific distance $\frac{\sqrt{3}}{2}a$ was carefully chosen by the examiner to make the subtended angle exactly $60^\circ$. If the distance changes, the angle changes, and so does the fraction of the flux.

Q: What if the rectangle was not centered at the origin?

If it were shifted along the x-axis, the angles subtended by the two halves would be different. However, you could still find the total angle subtended by the new edges and use the exact same symmetry argument to find the flux fraction.

Q: Is the flux through the flat rectangle really the same as the flux through a curved cylindrical arc?

Yes! As long as the flat surface and the curved surface subtend the exact same angle at the line charge and don't enclose any other charges between them, the flux through them is identical. This is because electric field lines are continuous and don't break.

Question

An infinitely long uniform line charge distribution of charge per unit length $\lambda$ lies parallel to the $y$-axis in the $y$-$z$ plane at $z = \frac{\sqrt{3}}{2}a$ (see figure). If the magnitude of the flux of the electric field through the rectangular surface $ABCD$ lying in the $x$-$y$ plane with its centre at the origin is $\frac{\lambda L}{n\varepsilon_0}$ ($\varepsilon_0 =$ permittivity of free space), then the value of $n$ is

Sigma Percentile · Accepted Answer

### Visualizing the Geometry

When faced with a three-dimensional electromagnetism problem, our first instinct is often to set up a double integral to calculate the electric flux. However, the true beauty of physics lies in recognizing symmetries that can turn a seemingly complex calculus problem into a straightforward geometry puzzle.

Imagine you are standing on the $y$-axis, looking directly at the $x$-$z$ plane. From this vantage point, the infinitely long line charge, which runs parallel to the $y$-axis, collapses into a single point located at $z = \frac{\sqrt{3}}{2}a$. Similarly, the rectangular surface $ABCD$, which lies flat in the $x$-$y$ plane, appears simply as a horizontal line segment of width $a$ resting on the $x$-axis.

By shifting our perspective, we have effectively reduced a daunting 3D problem into a much more manageable 2D cross-sectional view.

### The Power of Symmetry

An infinitely long uniform line charge produces an electric field that radiates outward uniformly in all radial directions. Because of this perfect cylindrical symmetry, the electric flux passing through any surface depends entirely on the angle that the surface subtends at the line charge.

Think of the line charge as a light bulb emitting rays equally in all directions. The amount of light hitting a screen depends on how much of the bulb's "field of view" the screen occupies. In our case, the flux through the flat rectangular surface is exactly the same as the flux through a curved cylindrical arc that subtends the exact same angle.

### Calculating the Subtended Angle

To find this crucial angle, we draw imaginary lines from the line charge down to the edges of our rectangle (which are at $x = -\frac{a}{2}$ and $x = \frac{a}{2}$). This forms an isosceles triangle, which the $z$-axis neatly bisects into two identical right-angled triangles.

Let's focus on one of these right-angled triangles. The base of this triangle is half the width of the rectangle, so $	ext{Base} = \frac{a}{2}$. The height of the triangle is the distance from the origin to the line charge, which is given as $	ext{Height} = \frac{\sqrt{3}}{2}a$.

Using basic trigonometry, we can find the half-angle $	heta$:

$$ 	an 	heta = \frac{	ext{Base}}{	ext{Height}} = \frac{a/2}{\frac{\sqrt{3}}{2}a} = \frac{1}{\sqrt{3}} $$

We know that the angle whose tangent is $\frac{1}{\sqrt{3}}$ is $30^\circ$. Therefore, the half-angle is $30^\circ$, which means the total angle subtended by the entire rectangle at the line charge is:

$$ 	ext{Total Angle} = 2 	imes 30^\circ = 60^\circ $$

### Fractional Flux and Final Answer

Now, we bring in Gauss's Law. If we were to enclose the line charge completely with a full cylinder of length $L$, it would subtend a full $360^\circ$. The total flux passing through this complete cylinder would be:

$$ \Phi_{	ext{total}} = \frac{q_{	ext{enclosed}}}{\varepsilon_0} = \frac{\lambda L}{\varepsilon_0} $$

Our rectangular surface doesn't surround the charge completely; it only intercepts a $60^\circ$ slice of the total $360^\circ$ field. Because the electric field is uniform in all radial directions, the flux through our rectangle is simply the proportional fraction of the total flux:

$$ \Phi_{	ext{rect}} = \left( \frac{60^\circ}{360^\circ} ight) \Phi_{	ext{total}} = \frac{1}{6} \left( \frac{\lambda L}{\varepsilon_0} ight) = \frac{\lambda L}{6\varepsilon_0} $$

The problem states that the flux is $\frac{\lambda L}{n\varepsilon_0}$. By comparing our derived expression with the given one, it becomes immediately clear that:

$$ n = 6 $$

This problem is a classic example of how exploiting symmetry and shifting your geometric perspective can bypass tedious integration, leading to an elegant and satisfying solution.

Enter Numerical Value:

Visualized Solution

$Cross-sectional View$

$Symmetry and Gauss’s Law$

$Subtended Angle Setup$

$Calculating the Half-Angle$

$Total Subtended Angle$

$Total Flux of Line Charge$

$Flux Through Rectangle$

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

Visualizing the Geometry

The Power of Symmetry

Calculating the Subtended Angle

Fractional Flux and Final Answer

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cross-sectional View

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Symmetry and Gauss’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Subtended Angle Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Half-Angle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Subtended Angle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Flux of Line Charge

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Flux Through Rectangle