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Chapter 11: Problem 1534

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

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

Step by step solution

01

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.
02

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: \[F_1 = k\frac{q \cdot q_0}{(a^2 + y^2)}\] \[F_2 = k\frac{q \cdot q_0}{(a^2 + y^2)}\]
03

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.
04

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).

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Most popular questions from this chapter

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