What will be the magnitude of electric field at point O as shown in figure? Each side of the figure

Q: Why do the fields from A and H cancel out?

Both A and H have a charge of $-q$ and are at the same distance from the center $O$. The electric field from a negative charge points towards it. Since A and H are on opposite sides of $O$ along the same diagonal, their fields point in exactly opposite directions and have the same magnitude, thus canceling each other out.

Q: Why is the distance for the diagonal charges $\sqrt{2}l$?

If you look at the grid, a diagonal charge like D is at coordinates $(l, l)$ relative to the center $(0,0)$. Using the Pythagorean theorem, the distance is $\sqrt{l^2 + l^2} = \sqrt{2l^2} = \sqrt{2}l$.

Q: Why did we subtract the fields on the x-axis?

Electric field is a vector. The charge at B pushes a positive test charge to the right, while the charge at G pushes it to the left. Since they point in opposite directions, we subtract their magnitudes to find the net field along that axis.

Q: What if the charges on the diagonals weren't symmetric?

If they weren't symmetric, we couldn't just cancel them. We would have to calculate the x and y components of the field from EACH diagonal charge using sine and cosine of $45^\circ$, and add them to the total x and y components. It would be much more math!

Q: Why is the final answer multiplied by $\sqrt{2}$?

The net field has an x-component of $E_0$ and a y-component of $E_0$ (where $E_0 = \frac{kq}{l^2}$). When you add two perpendicular vectors of equal magnitude, the resultant magnitude is $\sqrt{E_0^2 + E_0^2} = \sqrt{2E_0^2} = \sqrt{2}E_0$.

Question

What will be the magnitude of electric field at point $O$ as shown in figure? Each side of the figure is $l$ and perpendicular to each other.

Options:

(A) $\frac{1}{4\pi\varepsilon_0}\frac{q}{l^2}$

(B) $\frac{1}{4\pi\varepsilon_0}\frac{q}{l^2}(2\sqrt{2}-1)$

(C) $\frac{q}{4\pi\varepsilon_0(2l)^2}$

(D) $\frac{1}{4\pi\varepsilon_0}\frac{q}{l^2}(\sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are standing at the center $O$ of this intricate grid of charges. You are surrounded by eight distinct point charges, each exerting its own electric field on you. The problem asks us to find the net electric field at this exact center point.

At first glance, this looks like a nightmare of vector addition. We have charges on the x-axis, the y-axis, and even on the diagonals. But before we blindly start plugging numbers into Coulomb's law, we must pause and look for patterns. Physics rewards those who observe carefully.

### The Master Equation
The foundational tool we need is the formula for the electric field produced by a point charge:

$$E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$$

Since we are dealing with multiple charges, the principle of superposition tells us that the total electric field at $O$ is simply the vector sum of the individual fields from all eight charges.

### The Magic of Symmetry
Let's look at the diagonals first. Notice the charge at $A(-q)$ and the charge at $H(-q)$. They are located at opposite ends of the same diagonal, equidistant from the center $O$. Because they are both negative, their electric fields at $O$ point directly towards them. This means $\vec{E}_A$ and $\vec{E}_H$ are pulling in exactly opposite directions with the exact same strength. What happens? They perfectly cancel each other out!

$$\vec{E}_A + \vec{E}_H = 0$$

The exact same beautiful symmetry occurs on the other diagonal. The charges at $D(+q)$ and $E(+q)$ are both positive and equidistant from $O$. Their fields push away from them, directly into each other, resulting in another perfect cancellation.

$$\vec{E}_D + \vec{E}_E = 0$$

Just by observing symmetry, we have eliminated half of the charges from our calculation!

### Conquering the Axes
Now the problem is vastly simplified. Let's evaluate the horizontal (x) axis. We have a charge $B(+q)$ at a distance $l$ to the left, pushing to the right with a field $E_B = \frac{kq}{l^2}$. On the right, we have $G(2q)$ at a distance $l$, pushing to the left with a field $E_G = \frac{2kq}{l^2}$.

The leftward push is twice as strong, so the net field along the x-axis is:

$$E_x = E_G - E_B = \frac{2kq}{l^2} - \frac{kq}{l^2} = \frac{kq}{l^2} 	ext{ (towards left)}$$

We apply the same logic to the vertical (y) axis. At the top, $C(2q)$ pushes down with $E_C = \frac{2kq}{l^2}$. At the bottom, $F(q)$ pushes up with $E_F = \frac{kq}{l^2}$. The downward push dominates:

$$E_y = E_C - E_F = \frac{2kq}{l^2} - \frac{kq}{l^2} = \frac{kq}{l^2} 	ext{ (towards bottom)}$$

### Final Calculation
We are left with two perpendicular vectors: $E_x$ pointing left and $E_y$ pointing down, both having a magnitude of $\frac{kq}{l^2}$. To find the final resultant vector, we use the Pythagorean theorem:

$$E_{net} = \sqrt{E_x^2 + E_y^2}$$

$$E_{net} = \sqrt{\left(\frac{kq}{l^2}ight)^2 + \left(\frac{kq}{l^2}ight)^2} = \sqrt{2}\frac{kq}{l^2}$$

Finally, substituting $k = \frac{1}{4\pi\varepsilon_0}$, we arrive at our elegant final answer:

**$$E_{net} = \frac{1}{4\pi\varepsilon_0} \frac{q}{l^2} (\sqrt{2})$$**

This perfectly matches option (d). Always remember: symmetry is your greatest ally in physics!

Animated Solution for Physics - Electrostatics: What will be the magnitude of electric field at point $O$ as shown in figure? Each side of the figure is $l$ and perpendicular to each other.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Electric Field Formula$

$Cancellation on Diagonals$

$Field along X-axis$

$Field along Y-axis$

$Net Electric Field$

$Final Answer$

The Sigma Insight: Electric Field

Analyzing the Setup Imagine you are standing at the center $O$ of this intricate grid of charges

The Master Equation

The Magic of Symmetry Let's look at the diagonals first

Conquering the Axes Now the problem is vastly simplified

Final Calculation We are left with two perpendicular vectors: $E_{x}$ pointing left and $E_{y}$ pointing down, both having a magnitude of $\frac{k q}{l ^{2}}$

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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 Field Formula

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

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

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

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

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