Charges Q1 and Q2 are at points A and B of a right angle triangle OAB (see figure). The resultant el

Q: Why is the angle between $E_1$ and $E_{net}$ exactly $\theta$?

This is a classic geometry trick! The angle $\theta$ is inside the triangle between the hypotenuse $AB$ and the base $OB$. The field $E_1$ is parallel to the height $OA$ (both are vertical), and $E_{net}$ is perpendicular to the hypotenuse $AB$. Since the arms of the two angles are mutually perpendicular, the angles themselves must be equal.

Q: Why did we use $\tan \theta$ instead of $\sin$ or $\cos$?

We used $\tan \theta$ because it directly relates the two perpendicular components we know: $E_1$ and $E_2$. Using $\sin$ or $\cos$ would require calculating the magnitude of the resultant field $E_{net}$, which adds unnecessary algebraic complexity.

Q: Does it matter if the charges are positive or negative?

The problem implies they have the same sign because the resultant field points outward into the third quadrant to be perpendicular to the hypotenuse. If they had opposite signs, the resultant field would point into the second or fourth quadrant, and the geometry would change slightly, but the magnitude ratio would remain the same.

Q: What if the triangle wasn't a right-angled triangle?

If the triangle wasn't right-angled at $O$, the fields $E_1$ and $E_2$ would not be perpendicular to each other. We would have to use vector addition with the cosine rule to find the resultant, making the math significantly more involved.

Q: Can this be solved using vector dot products?

Absolutely! You can define the vector $\vec{AB} = x_2 \hat{i} - x_1 \hat{j}$ and the resultant field vector $\vec{E}_{net} = -E_2 \hat{i} - E_1 \hat{j}$. Since they are perpendicular, their dot product is zero: $\vec{AB} \cdot \vec{E}_{net} = 0$. This gives $-x_2 E_2 + x_1 E_1 = 0$, which leads to the exact same ratio!

Question

Charges $Q_1$ and $Q_2$ are at points $A$ and $B$ of a right angle triangle $OAB$ (see figure). The resultant electric field at point $O$ is perpendicular to the hypotenuse, then $Q_1/Q_2$ is proportional to

Options:

(A) $\frac{x_1^3}{x_2^3}$

(B) $\frac{x_2}{x_1}$

(C) $\frac{x_1}{x_2}$

(D) $\frac{x_2^2}{x_1^2}$

Sigma Percentile · Accepted Answer

The Harmony of Geometry and Electrostatics

Physics is often at its most beautiful when it perfectly intertwines with pure geometry. This problem is a classic example of how a physical constraint—the direction of a resultant electric field—can be elegantly unraveled using the properties of a right-angled triangle.

### Analyzing the Setup

Imagine a right-angled triangle $OAB$, with the right angle situated at the origin $O$. We place two point charges, $Q_1$ and $Q_2$, at vertices $A$ and $B$ respectively. The lengths of the sides adjacent to the right angle are given as $OA = x_1$ and $OB = x_2$.

Because both charges are positive, they generate electric fields that point away from themselves. At the origin $O$, the charge $Q_1$ (located on the y-axis) creates a downward electric field, which we will call $E_1$. Similarly, the charge $Q_2$ (located on the x-axis) creates a leftward electric field, $E_2$.

### The Physics: Coulomb's Law in Action

Using Coulomb's Law, we can easily write down the magnitudes of these two electric fields:

$$E_1 = \frac{k Q_1}{x_1^2}$$

$$E_2 = \frac{k Q_2}{x_2^2}$$

These two fields are perpendicular to each other, and they combine to form a resultant electric field, $\vec{E}_{	ext{net}}$.

### The Geometry: The Power of Perpendiculars

The problem provides a crucial piece of information: the resultant electric field $\vec{E}_{	ext{net}}$ is exactly perpendicular to the hypotenuse $AB$. This is where the magic happens.

Let's define the angle at vertex $B$ as $	heta$, so $\angle OBA = 	heta$. Now, consider the angle between the downward field $E_1$ and the resultant field $\vec{E}_{	ext{net}}$. By a fundamental theorem of geometry, if two lines are respectively perpendicular to two other lines, the angle between the first pair is equal to the angle between the second pair. Since $E_1$ is perpendicular to $OB$ (it lies along $OA$), and $\vec{E}_{	ext{net}}$ is perpendicular to $AB$, the angle between $E_1$ and $\vec{E}_{	ext{net}}$ must also be exactly $	heta$!

### The Master Equation

Now we can express $	an 	heta$ in two completely different ways.

First, using the vector components of the electric field:

$$	an 	heta = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{E_2}{E_1}$$

Substituting our expressions from Coulomb's Law:

$$	an 	heta = \frac{\frac{k Q_2}{x_2^2}}{\frac{k Q_1}{x_1^2}} = \frac{Q_2 x_1^2}{Q_1 x_2^2}$$

Second, we can find $	an 	heta$ directly from the spatial dimensions of the original right-angled triangle $OAB$:

$$	an 	heta = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{OA}{OB} = \frac{x_1}{x_2}$$

### Final Calculation

Since both expressions equal $	an 	heta$, we can equate them:

$$\frac{Q_2 x_1^2}{Q_1 x_2^2} = \frac{x_1}{x_2}$$

To find the ratio $Q_1 / Q_2$, we simply rearrange the terms. Notice how one power of $x_1$ and $x_2$ beautifully cancels out from both sides:

$$\frac{Q_2 x_1}{Q_1 x_2} = 1$$

$$\frac{Q_1}{Q_2} = \frac{x_1}{x_2}$$

And there we have it! The ratio of the charges is directly proportional to the ratio of their respective distances from the origin. A complex-looking vector problem melts away into a simple, elegant geometric proportion.

Animated Solution for Physics - Electrostatics: 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

Select Answer:

Visualized Solution

System Setup

Electric Fields at $O$

Resultant Field Condition

Angle Relationships

Evaluating $tan θ$ (Physics)

Evaluating $tan θ$ (Geometry)

Equating and Solving

Conclusion

The Sigma Insight: Electric Field

Analyzing the Setup

The Physics

The Geometry

The Master Equation

Final Calculation

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