Question

Two point positive charges \(q\) each are placed at \((-a, 0)\) and \((a, 0)\). A third positive charge \(q_{0}\) is placed at \((0, y)\). For which value of \(\mathrm{y}\) the force at \(q_{0}\) is maximum \(\ldots \ldots \ldots\) (A) a (B) \(2 \mathrm{a}\) (C) \((\mathrm{a} / \sqrt{2})\) (D) \((\mathrm{a} / \sqrt{3})\)

Short answer

The force at $q_0$ is maximum when \(y = \frac{a}{\sqrt{3}}\). Thus, the correct answer is (D).

Step-by-step solution

Write down Coulomb's Law for each charge

Coulomb's Law states that the electrostatic force between two point charges is given by \[F = k\frac{q_{1}q_{2}}{r^2}\] where F is the force between the charges, q1 and q2 are the magnitudes of the charges, r is the distance between the charges, and k is the electrostatic constant. We will find the forces acting on q0 due to the charges at (-a, 0) and (a, 0). Let F1 and F2 be the forces on q0 due to the charges at (-a, 0) and (a, 0), respectively.

Find the distance between charges and calculate forces

The distance between the charges q0 and q at (-a, 0) is \(\sqrt{a^2 + y^2}\) and between q0 and q at (a, 0) is also \(\sqrt{a^2 + y^2}\). Now, we can find F1 and F2 using Coulomb's Law: \[F_1 = k\frac{q \cdot q_0}{(a^2 + y^2)}\] \[F_2 = k\frac{q \cdot q_0}{(a^2 + y^2)}\]

Find the net force on q0 and its components

To find the net force on q0, we need to find the components of F1 and F2 along and perpendicular to the y-axis. Let's denote the angle between the F1 or F2 and the y-axis as θ. Then: \(\theta = \tan^{-1}(\frac{a}{y})\) Now, the y-component and x-component of each force can be found: \(F_{1y} = F_1 \cos{\theta}\) \(F_{1x} = F_1 \sin{\theta}\) \(F_{2y} = F_2 \cos{\theta}\) \(F_{2x} = -F_2 \sin{\theta}\) Now let's find the net force, Fnet, on q0: \(F_{net_y} = F_{1y} + F_{2y} = (F_1 + F_2) \cos{\theta}\) \(F_{net_x} = F_{1x} + F_{2x} = 0\) Since F1 and F2 are equal, and their x-component is equal and opposite, the net force on q0 is along the y-direction. So, Fnet = Fnet_y.

Find the condition for maximum net force

The net force on q0 is given by the y-component of the resultant force: \(F_{net} = (F_1+F_2)\cos{\theta}\) Let's differentiate Fnet with respect to y and set the derivative equal to zero to find the maximum: \(\frac{dF_{net}}{dy}=0\) We get: \[\frac{d((F_1+F_2)\cos{\theta})}{dy}=0\] Solving for y, we obtain: \(y = \frac{a}{\sqrt{3}}\) Therefore, the force at q0 is maximum when y = a/√3. So, the correct answer is (D).

Key concepts

Electrostatic Force

When we talk about electrostatic force, we're referring to the force that holds charged particles together. According to Coulomb's Law, the electrostatic force between two point charges is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This can be mathematically expressed as:

• \( F = k \frac{q_1 q_2}{r^2} \)

Where:

• \( F \) is the force between the charges
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges
• \( r \) is the distance between the charges
• \( k \) is the electrostatic constant

In the original exercise, we use Coulomb's Law to determine the forces acting upon a test charge due to two other charges positioned symmetrically along the x-axis.
The clever part of Coulomb's Law is how it enables us to calculate the force based solely on the properties of the charges and their distance. The symmetry in the given problem simplifies calculations, as both charges exert forces of the same magnitude on the third charge.

Components of Force

To find the net electrostatic force on the third charge in our scenario, it's essential to understand that we must consider the components of forces involved. When a single force acts at an angle to a reference axis, we break it down into two components: one along the axis and one perpendicular to it.
For forces \( F_1 \) and \( F_2 \) from our two point charges, we're particularly interested in their components along the y-axis. Given the symmetrical placement of charges, each force will have:

• a y-component given by \( F_{1y} = F_1 \cos{\theta} \)
• an x-component like \( F_{1x} = F_1 \sin{\theta} \)

Because of the setup's symmetry:

• The x-components cancel each other out, i.e., \( F_{net_x} = 0 \)
• The net force is solely along the y-axis: \( F_{net} = (F_1 + F_2)\cos{\theta} \)

These components make it easier to deal with forces in vector form, allowing us to accurately predict the behavior of the charge under influence of other charges.

Maximum Force Condition

Determining the condition for maximizing the force exerted on the charge \( q_0 \) boils down to optimizing the y-component of the resultant force from \( q \) charges. The goal is to find a value of \( y \) that satisfies the condition for maximum force. Mathematically, you find this condition by setting the derivative of the net force with respect to \( y \) to zero:

• \( \frac{dF_{net}}{dy} = 0 \)

This is a standard process in calculus to find maxima or minima of functions.
Solving this derivative gives the critical point \( y = \frac{a}{\sqrt{3}} \) for which the force is maximized. Thus, when \( y \) equals \( \frac{a}{\sqrt{3}} \), the forces from the two point charges constructively interfere along the y-direction, resulting in the maximum net force on the charge \( q_0 \). This optimization is key in many physics problems, allowing for determination of optimal parameters in various scenarios. Through this, you grasp not only the mathematical mechanics but also the elegance of interplay between geometry and force interactions.