The electric field in a region is given E = (3/5 E_0 i + 4/5 E_0 j) N/C. The ratio of flux of report

Q: Why does a surface parallel to the YZ-plane have an area vector along the X-axis?

The area vector is always defined as perpendicular (normal) to the surface. If a surface lies flat in the YZ-plane, the only direction perpendicular to both the Y and Z axes is the X-axis.

Q: What happened to the Z-component of the electric field?

The given electric field $\mathbf{E} = \frac{3}{5}E_0 \hat{\mathbf{i}} + \frac{4}{5}E_0 \hat{\mathbf{j}}$ has no $\hat{\mathbf{k}}$ component. This means the electric field is entirely confined to the XY-plane and does not point in the Z-direction at all.

Q: Why do we use the dot product to calculate flux?

Electric flux measures how much of the electric field is passing *through* a surface. The dot product $\mathbf{E} \cdot \mathbf{A} = EA \cos\theta$ perfectly captures this, as it multiplies the area by the component of the electric field that is perpendicular to the surface.

Q: Could the ratio be $2:4$ instead of $1:2$?

Mathematically, $1:2$ and $2:4$ represent the same ratio. However, in standard ratio notation $a:b$, we always reduce the numbers to their simplest integer form. Therefore, $1:2$ is the correct representation, making $a=1$.

Q: What if the electric field was not uniform?

If the electric field varied across the surface, we couldn't use the simple dot product $\mathbf{E} \cdot \mathbf{A}$. Instead, we would have to divide the surface into infinitesimally small areas $d\mathbf{A}$ and integrate the flux over the entire surface: $\phi = \int \mathbf{E} \cdot d\mathbf{A}$.

Question

The electric field in a region is given $\mathbf{E} = \left(\frac{3}{5}E_0 \hat{\mathbf{i}} + \frac{4}{5}E_0 \hat{\mathbf{j}}ight) \frac{	ext{N}}{	ext{C}}$. The ratio of flux of reported field through the rectangular surface of area $0.2 	ext{ m}^2$ (parallel to YZ-plane) to that of the surface of area $0.3 	ext{ m}^2$ (parallel to XZ- plane) is $a:b$, where $a = \dots\dots\dots$ .
[Here $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$ are unit vectors along X, Y and Z-axes, respectively]

Sigma Percentile · Accepted Answer

### Visualizing the Field and Surfaces

Imagine a region of space permeated by a uniform electric field. The problem gives us the electric field vector as $\mathbf{E} = \left(\frac{3}{5}E_0 \hat{\mathbf{i}} + \frac{4}{5}E_0 \hat{\mathbf{j}}ight) \frac{	ext{N}}{	ext{C}}$. This tells us that the field has components along both the X and Y axes, but no component along the Z axis.

We are asked to find the electric flux through two distinct rectangular surfaces. The first surface has an area of $0.2 	ext{ m}^2$ and is parallel to the YZ-plane. The second surface has an area of $0.3 	ext{ m}^2$ and is parallel to the XZ-plane.

### The Power of the Dot Product

To solve this, we need to recall the fundamental definition of electric flux. Electric flux, denoted by $\phi$, is a measure of the number of electric field lines passing through a given surface area. Mathematically, it is defined as the dot product of the electric field vector $\mathbf{E}$ and the area vector $\mathbf{A}$:

$$ \phi = \mathbf{E} \cdot \mathbf{A} $$

The area vector $\mathbf{A}$ is a vector whose magnitude is the area of the surface and whose direction is perpendicular (normal) to the surface.

### Calculating the Fluxes

Let's analyze the first surface. Since it is parallel to the YZ-plane, its normal vector must point along the X-axis. Therefore, the area vector for the first surface is:

$$ \mathbf{A}_1 = 0.2 \hat{\mathbf{i}} 	ext{ m}^2 $$

Now, we can calculate the flux $\phi_1$ through this surface by taking the dot product:

$$ \phi_1 = \mathbf{E} \cdot \mathbf{A}_1 = \left(\frac{3}{5}E_0 \hat{\mathbf{i}} + \frac{4}{5}E_0 \hat{\mathbf{j}}ight) \cdot (0.2 \hat{\mathbf{i}}) $$

Because the dot product of orthogonal unit vectors is zero ($\hat{\mathbf{j}} \cdot \hat{\mathbf{i}} = 0$) and parallel unit vectors is one ($\hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = 1$), we only multiply the X-components:

$$ \phi_1 = \frac{3}{5}E_0 	imes 0.2 = 0.12 E_0 $$

Next, we look at the second surface. It is parallel to the XZ-plane, which means its normal vector points along the Y-axis. Its area vector is:

$$ \mathbf{A}_2 = 0.3 \hat{\mathbf{j}} 	ext{ m}^2 $$

Calculating the flux $\phi_2$ through the second surface:

$$ \phi_2 = \mathbf{E} \cdot \mathbf{A}_2 = \left(\frac{3}{5}E_0 \hat{\mathbf{i}} + \frac{4}{5}E_0 \hat{\mathbf{j}}ight) \cdot (0.3 \hat{\mathbf{j}}) $$

This time, we only multiply the Y-components:

$$ \phi_2 = \frac{4}{5}E_0 	imes 0.3 = 0.24 E_0 $$

### The Final Ratio

Finally, we need to find the ratio of the flux through the first surface to the flux through the second surface:

$$ 	ext{Ratio} = \frac{\phi_1}{\phi_2} = \frac{0.12 E_0}{0.24 E_0} = \frac{1}{2} $$

The problem states that this ratio is $a:b$, and asks for the value of $a$. From our result, the ratio is $1:2$, which means $a = 1$ and $b = 2$.

**Therefore, the value of $a$ is 1.**

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Electric Flux Formula$

$Area Vector for Surface 1$

$Flux through Surface 1$

$Area Vector for Surface 2$

$Flux through Surface 2$

$Calculating the Ratio$

$Conclusion$

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

Visualizing the Field and Surfaces

The Power of the Dot Product

Calculating the Fluxes

The Final Ratio

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; } Visualizing the Setup

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Area Vector for Surface 1

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Area Vector for Surface 2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Flux through Surface 2

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

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