There are two large parallel metallic plates S1 and S2 carrying surface charge densities sigma1 and

Q: Why is the electric field of a metallic plate $\sigma/\varepsilon_0$ and not $\sigma/2\varepsilon_0$?

For a single non-conducting sheet of charge, the field is $\sigma/2\varepsilon_0$. However, a metallic plate has two surfaces. If $\sigma$ represents the charge density on the inner surface facing the other plate, the field is $\sigma/\varepsilon_0$. The problem implies this convention.

Q: Why do we use $\cos(\pi/4)$ for the work done?

The electric field is directed along the normal to the plates. The charge moves at an angle of $\pi/4$ to this normal. The work done is the dot product of force and displacement, which introduces the cosine of the angle between them.

Q: Does the charge's path matter for the work done?

No, the electrostatic force is conservative. The work done depends only on the initial and final positions, specifically the displacement parallel to the electric field lines.

Q: What if the charge was negative?

If the charge was negative, the force would be in the opposite direction to the electric field, and the work done by the electric field would be negative.

Q: Why is the net electric field $E_1 - E_2$?

Both plates produce electric fields. Since $\sigma_1$ and $\sigma_2$ are positive, their fields point away from them. In the region between the plates, these fields point in opposite directions, so we subtract them.

Question

There are two large parallel metallic plates $S_1$ and $S_2$ carrying surface charge densities $\sigma_1$ and $\sigma_2$ respectively ($\sigma_1 > \sigma_2$) placed at a distance $d$ apart in vacuum. Find the work done by the electric field in moving a point charge $q$ a distance $a (a < d)$ from $S_1$ towards $S_2$ along a line making an angle $\pi / 4$ with the normal to the plates.

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are standing between two massive, parallel metallic walls, $S_1$ and $S_2$. These aren't just any walls; they are charged! Plate $S_1$ carries a surface charge density of $\sigma_1$, and plate $S_2$ carries $\sigma_2$. We are given that $\sigma_1$ is greater than $\sigma_2$.

Because these are large metallic plates, the electric field produced by each plate in the region between them is given by the formula $E = \frac{\sigma}{\varepsilon_0}$. Since both plates are positively charged, their electric fields point away from them. In the space between the plates, these two fields oppose each other.

### The Master Equation for the Net Field
Since $\sigma_1 > \sigma_2$, the electric field from $S_1$ overpowers the field from $S_2$. The net electric field, $\vec{E}_{	ext{net}}$, will point directly from $S_1$ towards $S_2$.

We can calculate its magnitude by simply subtracting the weaker field from the stronger one:
$$E_{	ext{net}} = E_1 - E_2 = \frac{\sigma_1}{\varepsilon_0} - \frac{\sigma_2}{\varepsilon_0} = \frac{\sigma_1 - \sigma_2}{\varepsilon_0}$$

### Calculating the Work Done
Now, a point charge $q$ enters the scene. It moves a distance $a$ from $S_1$ towards $S_2$. However, it doesn't take the shortest path straight across. Instead, it moves along a slanted path, making an angle of $	heta = \frac{\pi}{4}$ with the normal (the straight line connecting the plates).

We need to find the work done by the electric field on this charge. The fundamental definition of work done by a constant force is the dot product of the force vector and the displacement vector:
$$W = \vec{F} \cdot \vec{s} = q\vec{E}_{	ext{net}} \cdot \vec{s}$$

This dot product expands to:
$$W = q E_{	ext{net}} s \cos 	heta$$

### Final Calculation
Let's plug in all the pieces of our puzzle. The displacement magnitude $s$ is $a$, the angle $	heta$ is $\frac{\pi}{4}$, and we have our expression for $E_{	ext{net}}$.

$$W = q \left( \frac{\sigma_1 - \sigma_2}{\varepsilon_0} ight) (a) \cos \left( \frac{\pi}{4} ight)$$

We know from basic trigonometry that $\cos \left( \frac{\pi}{4} ight) = \frac{1}{\sqrt{2}}$. Substituting this in gives us our final, elegant result:

**$$W = \frac{(\sigma_1 - \sigma_2)qa}{\sqrt{2}\varepsilon_0}$$**

This result beautifully illustrates a core principle of physics: when a force is conservative like the electrostatic force, only the component of displacement parallel to the force field actually contributes to the work done!

Visualized Solution

Visualizing the Setup

Net Electric Field

Path of the Charge

Work Done Formula

Substituting Values

Final Calculation

Conclusion

The Sigma Insight: Electric Potential and Potential Difference

Analyzing the Setup Imagine you are standing between two massive, parallel metallic walls, $S_{1}$ and $S_{2}$

The Master Equation for the Net Field Since $σ_{1} > σ_{2}$ , the electric field from $S_{1}$ overpowers the field from $S_{2}$

Calculating the Work Done Now, a point charge $q$ enters the scene

Final Calculation Let's plug in all the pieces of our puzzle

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