An infinite number of point charges, each carrying 1 μC charge, are placed along the Y-axis at y = 1

Q: Why do we add all the forces directly?

Because all the charges are positive and lie on the same side of the origin along the Y-axis. Therefore, all force vectors point in the exact same direction (negative Y-direction), allowing us to simply add their magnitudes.

Q: What if the charges were placed at $1, 3, 9, 27$ instead?

The process remains exactly the same, but your common ratio $r$ in the geometric progression would change. For distances $3^n$, the squared distances would be $9^n$, making the common ratio $1/9$.

Q: Can the sum of infinite forces be finite?

Yes! This is the beauty of convergent infinite series. Because the distance squared grows so rapidly, the force from distant charges becomes infinitesimally small, allowing the total sum to converge to a finite value.

Q: What if the charges had alternating signs?

If the charges were $+1\mu\text{C}, -1\mu\text{C}, +1\mu\text{C}$, the forces would alternate in direction. The series would still be a GP, but the common ratio $r$ would be negative ($-1/4$), leading to a different finite sum.

Q: Why did we convert microcoulombs to coulombs?

To ensure our final force is in standard SI units (Newtons). Always convert all units to the standard SI system before performing calculations to avoid massive power-of-ten errors.

Question

An infinite number of point charges, each carrying $1\mu	ext{C}$ charge, are placed along the Y-axis at $y = 1	ext{m}, 2	ext{m}, 4	ext{m}, 8	ext{m} \dots$. The total force on a $1	ext{ C}$ point charge, placed at the origin, is $x 	imes 10^3	ext{ N}$. The value of $x$ to the nearest integer, is ............... .
(Take, $\frac{1}{4\pi\varepsilon_0} = 9 	imes 10^9 	ext{ N-m}^2/	ext{C}^2$)

Sigma Percentile · Accepted Answer

The problem of infinite charges is a beautiful intersection where the physical laws of electrostatics meet the mathematical elegance of infinite series. It is a classic setup that tests not just your knowledge of Coulomb's Law, but your ability to recognize patterns.

### The Setup: Infinity on an Axis

Imagine standing at the origin of a coordinate system. Right where you stand, a massive $1	ext{ C}$ test charge is placed. Now, look up along the Y-axis. At $y = 1	ext{ m}$, there is a $1\mu	ext{C}$ charge. At $y = 2	ext{ m}$, there is another. Then at $4	ext{ m}$, $8	ext{ m}$, and so on, stretching out into infinity.

Because all these charges are positive, they all exert a repulsive force on the $1	ext{ C}$ charge at the origin, pushing it downwards in the negative Y-direction. To find the total force, we must invoke the **Principle of Superposition**, which states that the net force is simply the vector sum of all individual forces.

### The Superposition Principle

According to Coulomb's Law, the force between two charges is given by:
$$F = \frac{kQq}{r^2}$$

Let's write out the sum of the forces exerted by each charge on the Y-axis:
$$F_{	ext{net}} = \frac{kQq}{1^2} + \frac{kQq}{2^2} + \frac{kQq}{4^2} + \frac{kQq}{8^2} + \dots$$

At first glance, summing an infinite number of forces might seem impossible. But let's factor out the constants to see the underlying structure:
$$F_{	ext{net}} = kQq \left[ \frac{1}{1} + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots ight]$$

### The Mathematical Twist: Infinite GP

Look closely at the series inside the bracket. Each term is exactly one-fourth of the previous term. This is a classic **Infinite Geometric Progression (GP)**!

For an infinite GP, the sum converges to a finite value if the common ratio $r$ is strictly between $-1$ and $1$. Here, our first term is $a = 1$ and our common ratio is $r = \frac{1}{4}$. The formula for the sum of an infinite GP is:
$$S_\infty = \frac{a}{1 - r}$$

Substituting our values:
$$S_\infty = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$

This elegant result tells us that the infinite sequence of charges exerts a force equivalent to a single charge placed at a specific effective distance!

### The Final Calculation

Now, we bring physics back into the picture. We substitute the sum of the series and the known values of the constants back into our force equation. Remember to convert microcoulombs to coulombs!
$$F_{	ext{net}} = (9 	imes 10^9) 	imes (1) 	imes (1 	imes 10^{-6}) 	imes \left[ \frac{4}{3} ight]$$

Simplifying the powers of ten:
$$F_{	ext{net}} = 9 	imes 10^3 	imes \frac{4}{3}$$
$$F_{	ext{net}} = 12 	imes 10^3	ext{ N}$$

The problem states that the net force is $x 	imes 10^3	ext{ N}$. By directly comparing our result, we find the final answer:
**$x = 12$**

Enter Numerical Value:

Visualized Solution

Setup of Charges at $y = 2^{n}$

Superposition Principle

Substituting Values

Factoring Constants

Infinite Geometric Progression

Sum of Infinite GP

Final Substitution

Comparing with $x \times 1 0^{3}$

Alternating Charges

The Sigma Insight: Coulomb's Law

The Setup

The Superposition Principle

The Mathematical Twist

The Final Calculation

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