Problem 61 - Master Resource Book in JEE Main Physics

Question

A charge \(+q\) is fixed at each of the points \(x=x_{0}\) \(x=3 x_{0}, x=5 x_{0} \ldots \infty\), on the \(x\)-axis and a charge \(-q\) is fixed at each of the points \(x=2 x_{0}, x=4 x_{0} x=6 x_{0} \ldots \infty .\) Here, \(x_{0} \quad\) is the constant. Take the electric potential at a point due to a charge \(Q\) at a distance \(\mathrm{r}\) from it to be \(Q / 4 \pi \varepsilon_{0} r\). Then, the potential at the origin due to the above system of charges is (a) \(\frac{q}{4 \pi \varepsilon_{0} x_{0}} \log _{c} 2\) (b) \(\frac{q}{8 \pi \varepsilon_{0} x_{0}} \log _{c} 2\) (c) 0 (d) \(\infty\)

Step-by-Step Solution

Answer

The potential at the origin is \(\frac{q}{4 \pi \varepsilon_0 x_0} \log(2)\) (Option a).

Step 1: Identify the Charges and Positions

We have charges placed along the x-axis at specific intervals. Positive charges "+q" are positioned at odd multiples of \(x_0\) (i.e., \(x = x_0, 3x_0, 5x_0, ...\)) and negative charges "-q" at even multiples of \(x_0\) (i.e., \(x = 2x_0, 4x_0, 6x_0, ...\)).

Step 2: Determine the Potential Due to a Single Charge

The electric potential \(V\) at a point due to a single charge \(Q\) at distance \(r\) is given by \(V = \frac{Q}{4 \pi \varepsilon_0 r}\). This formula will be used to calculate the potential contribution of each charge at the origin \(x = 0\).

Step 3: Calculate Potentials for Positive Charges

Considering the positive charges at \(x = x_0, 3x_0, 5x_0, ...\), the potentials at the origin are \(\frac{+q}{4 \pi \varepsilon_0 x_0}\), \(\frac{+q}{4 \pi \varepsilon_0 3x_0}\), \(\frac{+q}{4 \pi \varepsilon_0 5x_0}\), and so on.

Step 4: Calculate Potentials for Negative Charges

For the negative charges at \(x = 2x_0, 4x_0, 6x_0, ...\), the potentials are \(\frac{-q}{4 \pi \varepsilon_0 2x_0}\), \(\frac{-q}{4 \pi \varepsilon_0 4x_0}\), \(\frac{-q}{4 \pi \varepsilon_0 6x_0}\), etc.

Step 5: Calculate the Total Potential at the Origin

Sum the potentials from the positive charges and the negative charges. The series for positive charges potential becomes \(\frac{q}{4 \pi \varepsilon_0 x_0} \sum_{n=0}^{ ext{even}} \frac{1}{2n+1}\), and for negative charges \(\frac{-q}{4 \pi \varepsilon_0 x_0} \sum_{n=1}^{ ext{odd}} \frac{1}{n}\). This simplifies to a known mathematical series.

Step 6: Simplify the Series Using Known Results

The series \(\frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots = \log(2)\). Using this, the total potential becomes \(\frac{q}{4 \pi \varepsilon_0 x_0} \log(2)\).

Step 7: Compare and Conclude

Compare the result \(\frac{q}{4 \pi \varepsilon_0 x_0} \log(2)\) with the provided options. The correct answer is option (a) \(\frac{q}{4 \pi \varepsilon_0 x_0} \log_{c} 2\).

Key Concepts

Electric Potential

Electric potential is a fundamental concept in electrostatics, which describes the potential energy per unit charge at a particular point in space resulting from electrostatic forces. It's usually measured in volts and denoted by the symbol \(V\).

The formula for the electric potential \(V\) due to a point charge \(Q\) at a distance \(r\) is given by:

• \( V = \frac{Q}{4\pi\varepsilon_0 r} \)

Here, \(\varepsilon_0\) is the permittivity of free space, a constant value which allows us to determine the effect of electric forces in a vacuum.

Understanding electric potential is crucial because it helps in calculating the work done in moving a charge within an electric field. It acts as a bridge to more complicated problems involving multiple charges, allowing us to add potentials algebraically when calculating the net potential at a point in space.

Point Charges

Point charges are idealized charges that are used to model the behavior of electric charges in space. They are considered to have no dimensions, meaning all the charge is concentrated at a single point. This simplifies the analysis of electric fields and potentials.

In this exercise, we have both positive and negative point charges arranged along the x-axis:

• Positive charges are located at odd multiples of \(x_0\) such as \(x = x_0, 3x_0, 5x_0, \cdots\).
• Negative charges are positioned at even multiples of \(x_0\) like \(x = 2x_0, 4x_0, 6x_0, \cdots\).

Each charge creates an electric potential at any point that is inversely proportional to its distance from that point, as noted in the electric potential formula. Despite being point charges, when arranged systematically, they can generate complex potential patterns that require detailed computation.

Being able to identify and understand how point charges contribute to electric fields and potentials is critical in numerous fields of physics and engineering.

Mathematical Series in Physics

Mathematical series play an essential role in solving physics problems, especially when dealing with sequences or sums of functions. In the study of electrostatics, series can determine the cumulative effect of an infinite number of charges.

A thorough understanding of series helps us in the context of multiple charges arranged systematically, such as in this exercise. For example, the potential at the origin is determined by summing series contributions from both positive and negative charges:

• Positive charge series: \( \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \ldots \)
• Negative charge series: \( \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots \)

To simplify these series, known mathematical results such as the alternating harmonic series can be employed:

• An example of such a simplification is \( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \cdots = \log(2) \).

Incorporating series into physics allows for solving infinite arrangements like this effectively, achieving a solution that aligns with known results while maintaining the integrity of the physical problem.