Three charges +Q, q, +Q are placed respectively at distance 0, d/2 and d from the origin on the X-ax

Q: Why do we only calculate the force on the charge at the origin?

Because the problem specifically states a condition for that particular charge: 'the net force experienced by $+Q$ placed at $x=0$ is zero'. We use this given condition to set up our equation.

Q: How do we know if the forces are attractive or repulsive before we know the sign of $q$?

We don't need to guess! If we assume $q$ is positive and write the sum of the forces as $F_q + F_Q = 0$, the math will naturally take care of the signs. Since $F_Q$ is definitely repulsive (positive), $F_q$ must turn out to be negative (attractive) for the sum to be zero. The math reveals that $q$ must be negative.

Q: Why did $(d/2)^2$ become $4/d^2$ in the numerator?

When you square a fraction, you square both the numerator and denominator: $(d/2)^2 = d^2 / 2^2 = d^2 / 4$. Dividing by a fraction is the same as multiplying by its reciprocal. So, $1 / (d^2 / 4)$ becomes $4 / d^2$.

Q: What if the middle charge was placed at $d/3$ instead of $d/2$?

The logic remains exactly the same, but the distance in the first term of your equation would be $d/3$. The equation would be $\frac{kQq}{(d/3)^2} + \frac{kQ^2}{d^2} = 0$, which simplifies to $9q + Q = 0$, giving $q = -Q/9$.

Q: Is the middle charge $q$ also in equilibrium?

Yes! Because it is placed exactly in the middle of two identical charges ($+Q$ and $+Q$), the forces from the left and right will be equal in magnitude and opposite in direction, perfectly canceling each other out regardless of the value of $q$.

Question

Three charges $+Q, q, +Q$ are placed respectively at distance $0, d/2$ and $d$ from the origin on the X-axis. If the net force experienced by $+Q$ placed at $x=0$ is zero, then value of $q$ is

Options:

(A) $+\frac{Q}{2}$

(B) $+\frac{Q}{4}$

(C) $-\frac{Q}{2}$

(D) $-\frac{Q}{4}$

Sigma Percentile · Accepted Answer

### Setting the Stage

Imagine a straight line, the $x$-axis, acting as a tightrope where three electrical charges are carefully placed. At the very beginning, the origin ($x = 0$), sits a positive charge $+Q$. Exactly halfway down the line at $x = d/2$, we place an unknown charge $q$. Finally, at the end of our segment at $x = d$, we place another positive charge $+Q$.

The problem presents us with a fascinating condition: the charge sitting at the origin is perfectly at peace. It experiences absolutely zero net force. Our mission is to uncover the identity—both the sign and the magnitude—of that mysterious middle charge $q$ that makes this perfect balance possible.

### The Principle of Superposition

In the world of electrostatics, charges don't just interact in pairs and ignore the rest; they all talk to each other simultaneously. The **Principle of Superposition** tells us that the total force on our target charge at the origin is simply the vector sum of the individual forces exerted by every other charge in the system.

So, the net force $\vec{F}_{net}$ on the charge at $x=0$ is the sum of the force from the middle charge $q$ and the force from the far charge $+Q$.

$$ \vec{F}_{net} = \vec{F}_{q} + \vec{F}_{Q} = 0 $$

### The Math Unfolds

To translate this physical reality into mathematics, we call upon **Coulomb's Law**, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them: $F = \frac{k q_1 q_2}{r^2}$.

Let's write down the forces acting on the charge at the origin.

First, the force from the middle charge $q$, which is at a distance of $d/2$:
$$ F_q = \frac{k \cdot Q \cdot q}{(d/2)^2} $$

Second, the force from the far charge $+Q$, which is at a distance of $d$:
$$ F_Q = \frac{k \cdot Q \cdot Q}{d^2} $$

Since the net force is zero, their sum must vanish:
$$ \frac{k \cdot Q \cdot q}{(d/2)^2} + \frac{k \cdot Q^2}{d^2} = 0 $$

Now, let's carefully expand that denominator in the first term. $(d/2)^2$ becomes $d^2/4$. When we bring that $4$ up to the numerator, our equation transforms into:
$$ \frac{4kQq}{d^2} + \frac{kQ^2}{d^2} = 0 $$

This looks much friendlier! Notice how both terms share several common factors: $k$, $Q$, and $d^2$. Since none of these are zero, we can safely divide the entire equation by $\frac{kQ}{d^2}$.

This elegant cancellation strips away the complexity, leaving us with a beautifully simple linear equation:
$$ 4q + Q = 0 $$

### The Grand Conclusion

Solving this final equation is a breeze. We simply move $Q$ to the other side and divide by $4$:
$$ 4q = -Q $$
$$ q = -\frac{Q}{4} $$

And there we have it! For the charge at the origin to feel no net push or pull, the middle charge must be negative (to provide an attractive force that perfectly cancels the repulsive force from the far $+Q$ charge) and its magnitude must be exactly one-quarter of $Q$.

This perfectly matches option (d).

As a thought experiment, try calculating the net force on the middle charge $q$ now that you know its value. You'll find that the entire system is in a state of delicate equilibrium!

Animated Solution for Physics - Electrostatics: Three charges $+ Q, q, + Q$ are placed respectively at distance $0, d /2$ and $d$ from the origin on the X-axis. If the net force experienced by $+ Q$ placed at $x = 0$ is zero, then value of $q$ is

Select Answer:

Visualized Solution

Visualizing the Setup

Principle of Superposition

Applying Coulomb's Law

Simplifying the Expression

Canceling Common Terms

Solving for $q$

Final Conclusion

The Way Forward

The Sigma Insight: Coulomb's Law

Setting the Stage

The Principle of Superposition

The Math Unfolds

The Grand Conclusion

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