Three charges Q, +q and +q are placed at the vertices of a right angle isosceles triangle as shown b

Q: Why do we have to consider pairs of charges instead of individual charges?

Electrostatic potential energy is fundamentally an interaction energy between two charges. A single isolated charge doesn't have potential energy in its own right. For a system, we must account for all unique pairwise interactions to find the total energy required to assemble the system.

Q: Does the sign of the charge actually matter in the energy formula?

Yes, absolutely! Potential energy is a scalar quantity, and the sign of the charges determines whether the energy is positive (which implies a repulsive system) or negative (which implies an attractive, bound system). Always plug in the charges with their signs.

Q: What if we had four charges instead of three? How many pairs would there be?

For four charges, you would have to consider 6 unique pairs. You can find this using the combinations formula $^nC_2$, which for $n=4$ gives $\frac{4 \times 3}{2} = 6$. You would sum the energies of all 6 pairs.

Q: Why did we assume the sides to be $a$? The problem didn't give a side length.

The exact side length doesn't matter because it factors out and cancels completely when we set the total energy to zero. We just need a dummy variable to represent the ratio of the distances correctly (like $a$ and $\sqrt{2}a$).

Q: Can the net electrostatic energy ever be zero if all the charges are positive?

No, it's mathematically impossible. If all charges are positive, all pairwise energies are positive, and their sum must be strictly greater than zero. At least one charge must be negative to create negative potential energy terms that can cancel out the positive ones.

Question

Three charges $Q$, $+q$ and $+q$ are placed at the vertices of a right angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of $Q$ is

Options:

(A) $-2q$

(B) $\frac{-q}{1+\sqrt{2}}$

(C) $+q$

(D) $\frac{-\sqrt{2}q}{\sqrt{2}+1}$

Sigma Percentile · Accepted Answer

### Visualizing the Geometry

Imagine you are looking at a right-angled isosceles triangle. Let's assume the two equal sides that form the right angle have a length of $a$. By applying the Pythagorean theorem, the hypotenuse will naturally have a length of $\sqrt{a^2 + a^2} = \sqrt{2}a$.

At the vertices of this triangle, we have three point charges. Two positive charges, $+q$, are placed at the ends of one of the sides of length $a$ (let's say the bottom-left and bottom-right vertices). The third charge, $Q$, is placed at the top vertex.

### The Principle of Electrostatic Potential Energy

To find the total electrostatic potential energy of any system of charges, we must sum up the potential energies of all possible unique pairs in the system. The potential energy between any two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by the standard formula:

$$U = \frac{k q_1 q_2}{r}$$

For our specific three-charge system, we will have exactly three unique pairs. We need to calculate the energy for each pair and add them together.

### Setting Up the Master Equation

Let's write down the energy for each of the three pairs:

1.  **The two $+q$ charges:** They are separated by a distance $a$. Their interaction energy is $\frac{k(+q)(+q)}{a} = \frac{kq^2}{a}$.
2.  **The charge $Q$ and the first $+q$ charge:** They are separated by the vertical side of length $a$. Their interaction energy is $\frac{k(Q)(+q)}{a}$.
3.  **The charge $Q$ and the second $+q$ charge:** They are separated by the hypotenuse of length $\sqrt{2}a$. Their interaction energy is $\frac{k(Q)(+q)}{\sqrt{2}a}$.

Summing these up gives us the total electrostatic energy of the configuration:

$$U = \frac{kq^2}{a} + \frac{kQq}{a} + \frac{kQq}{\sqrt{2}a}$$

### Applying the Zero Energy Constraint

The problem states a very crucial condition: the net electrostatic energy of this entire configuration is exactly zero. So, we equate our total energy expression to zero:

$$\frac{kq^2}{a} + \frac{kQq}{a} + \frac{kQq}{\sqrt{2}a} = 0$$

Now, notice that the term $\frac{kq}{a}$ is common in all three terms. Let's factor it out to simplify the equation:

$$\frac{kq}{a} \left( q + Q + \frac{Q}{\sqrt{2}} \right) = 0$$

### The Final Calculation

Since the electrostatic constant $k$, the distance $a$, and the charge $q$ are non-zero, the term inside the bracket must be zero for the equation to hold true.

$$q + Q \left( 1 + \frac{1}{\sqrt{2}} \right) = 0$$

Let's isolate $Q$. We can take $q$ to the other side, making it $-q$, and take a common denominator for the terms with $Q$:

$$Q \left( \frac{\sqrt{2}+1}{\sqrt{2}} \right) = -q$$

Finally, we cross-multiply to solve for **$Q$**:

**$$Q = \frac{-\sqrt{2}q}{\sqrt{2}+1}$$**

This perfectly matches option (d).

### A Thought Experiment

Think about this: what if the triangle was equilateral instead of right-angled? All distances between the charges would simply be $a$. The math would be even simpler! The equation would be $\frac{kq^2}{a} + \frac{2kQq}{a} = 0$, and $Q$ would come out to be $-\frac{q}{2}$. Always visualize the geometry before jumping into the equations; it builds strong physical intuition!

Animated Solution for Physics - Electrostatics: Three charges $Q$ , $+ q$ and $+ q$ are placed at the vertices of a right angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of $Q$ is

Select Answer:

Visualized Solution

System Geometry

Electrostatic Potential Energy

Energy of the Configuration

Applying the Given Condition

Factoring the Equation

Solving for $Q$

Final Answer

The Way Forward

The Sigma Insight: Electrostatic Potential Energy

Visualizing the Geometry

The Principle of Electrostatic Potential Energy

Setting Up the Master Equation

Applying the Zero Energy Constraint

The Final Calculation

A Thought Experiment

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