Three charged particles A, B and C with charges -4q, 2q and -2q are present on the circumference of

Q: Why does the electric field of charge B point towards -30 degrees?

Charge B is positive, so its electric field points *away* from it. Since B is located at an angle of $150^\circ$ (in the second quadrant), the direction pointing directly away from it passes through the origin and goes into the fourth quadrant, exactly at $150^\circ - 180^\circ = -30^\circ$.

Q: Why do we ignore the negative signs of the charges when calculating the magnitude?

The negative sign of a charge only determines the *direction* of the electric field (towards the charge). Once we use the sign to draw the vector in the correct direction, we only use the absolute value of the charge to calculate the magnitude.

Q: What happens to the y-components of the electric field?

If you calculate the y-components, you'll find that the upward y-component from charge A ($4E_0 \sin 30^\circ$) is exactly canceled out by the downward y-components from charges B and C ($2E_0 \sin 30^\circ + 2E_0 \sin 30^\circ$). So, the net field is purely along the x-axis!

Q: Why is the distance $r$ taken as $d$ for all charges?

The problem states that the charges are placed on the circumference of a circle of radius $d$, and we are calculating the field at the center $O$. The distance from the center to any point on the circumference is simply the radius, $d$.

Q: Can we solve this using vector notation (i, j) instead of resolving components?

Absolutely! You can write each position vector, find the unit vectors, and use $\vec{E} = \frac{kq}{r^2} \hat{r}$. However, resolving components geometrically using angles is often faster and less prone to algebraic errors in symmetric setups like this.

Question

Three charged particles $A, B$ and $C$ with charges $-4q, 2q$ and $-2q$ are present on the circumference of a circle of radius $d$. The charged particles $A, C$ and centre $O$ of the circle formed an equilateral triangle as shown in figure. Electric field at $O$ along x-direction is

Options:

(A) $\frac{2\sqrt{3}q}{\pi\varepsilon_0 d^2}$

(B) $\frac{\sqrt{3}q}{4\pi\varepsilon_0 d^2}$

(C) $\frac{3\sqrt{3}q}{4\pi\varepsilon_0 d^2}$

(D) $\frac{\sqrt{3}q}{\pi\varepsilon_0 d^2}$

Sigma Percentile · Accepted Answer

### The Dance of Electric Fields: Mastering Vector Superposition

Imagine you are standing at the center of a circle, surrounded by three charged particles. This is a classic electrostatics problem that tests your ability to visualize vectors and apply the principle of superposition. Let's break down the setup and see how these electric fields interact.

### Analyzing the Setup

We have a circle of radius $d$ with three charges placed on its circumference. 
- Charge $A$ is $-4q$ and is located at an angle of $30^\circ$.
- Charge $C$ is $-2q$ and is located at an angle of $-30^\circ$.
- Charge $B$ is $+2q$ and is located at an angle of $150^\circ$.

Our goal is to find the net electric field at the center $O$, specifically along the x-direction.

### The Master Equation: Coulomb's Law

To find the net electric field, we rely on Coulomb's law. The magnitude of the electric field due to a point charge is given by:

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

Remember, the direction is crucial. The electric field points **away** from positive charges and **towards** negative charges.

### Breaking Down the Vectors

Let's analyze the electric field produced by each charge individually at the center $O$.

**1. Field due to Charge A:**
Charge $A$ is negative ($-4q$), so its electric field $\vec{E}_A$ points directly towards $A$. The angle is $30^\circ$. Its magnitude is:

$E_A = \frac{k(4q)}{d^2}$

**2. Field due to Charge B:**
Charge $B$ is positive ($+2q$), so its electric field $\vec{E}_B$ points directly away from $B$. Since $B$ is at $150^\circ$, pointing away means the field is directed exactly towards $150^\circ - 180^\circ = -30^\circ$. Its magnitude is:

$E_B = \frac{k(2q)}{d^2}$

**3. Field due to Charge C:**
Charge $C$ is negative ($-2q$), so its electric field $\vec{E}_C$ points directly towards $C$, which is at $-30^\circ$. Its magnitude is:

$E_C = \frac{k(2q)}{d^2}$

### The Magic of Alignment

Did you notice something beautiful? The electric fields from charges $B$ and $C$ are pointing in the exact same direction! Both $\vec{E}_B$ and $\vec{E}_C$ are directed along the $-30^\circ$ line. This perfect alignment simplifies our calculation immensely.

### Resolving Components

The question specifically asks for the net electric field along the x-direction. We need to find the x-components of these three vectors. We do this by multiplying each magnitude by the cosine of its angle with the x-axis. Interestingly, the angle with the x-axis is $30^\circ$ for all of them (since $\cos(-30^\circ) = \cos(30^\circ)$).

$E_x = E_A \cos 30^\circ + E_B \cos 30^\circ + E_C \cos 30^\circ$

Let's substitute the magnitudes we found:

$E_x = \left( \frac{4kq}{d^2} + \frac{2kq}{d^2} + \frac{2kq}{d^2} \right) \cos 30^\circ$

### Final Calculation

Adding the terms inside the bracket gives us:

$E_x = \left( \frac{8kq}{d^2} \right) \cos 30^\circ$

Now, we substitute $k = \frac{1}{4\pi\varepsilon_0}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$:

$E_x = \frac{1}{4\pi\varepsilon_0} \cdot \frac{8q}{d^2} \cdot \frac{\sqrt{3}}{2}$

Simplifying this expression, we arrive at our final answer:

**$E_x = \frac{\sqrt{3}q}{\pi\varepsilon_0 d^2}$**

It's a brilliant problem that perfectly demonstrates the power of vector superposition and geometric symmetry. If you were to calculate the y-components, you would find that they perfectly cancel each other out, leaving a pure horizontal field!

Select Answer:

Visualized Solution

$V i s u a l i z in g t h e S e t u p$

$C o u l o m b^{'} s L a w$

$F i e l d d u e t o C ha r g e A$

$F i e l d d u e t o C ha r g e B$

$F i e l d d u e t o C ha r g e C$

$R eso l v in g C o m p o n e n t s a l o n g X - a x i s$

$S u b s t i t u t in g M a g ni t u d es$

$F ina l C a l c u l a t i o n$

$C o n c l u s i o n$

The Sigma Insight: Electric Field

The Dance of Electric Fields

Analyzing the Setup

The Master Equation

Breaking Down the Vectors

The Magic of Alignment

Resolving Components

Final Calculation

Similar Questions

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Charges $Q_{1}$ and $Q_{2}$ are at points $A$ and $B$ of a right angle triangle $O A B$ (see figure). The resultant electric field at point $O$ is perpendicular to the hypotenuse, then $Q_{1} / Q_{2}$ is proportional to

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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; } Coulomb′s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Field due to Charge A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Field due to Charge B

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Field due to Charge C

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

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

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

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