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Chapter 11: Problem 1562

A Charge \(q\) is placed at the centre of the open end of cylindrical vessel. The flux of the electric field through the surface of the vessel is $\ldots \ldots \ldots \ldots$ (A) \(\left(\mathrm{q} / \in_{0}\right)\) (B) (q / \(2 \in_{0}\) ) (C) \(\left(2 q / \epsilon_{0}\right)\) (D) Zero

Short Answer

Expert verified
The electric flux through the surface of the open cylindrical vessel is Φ_vessel = \( \frac{1}{2} \) \( \frac{q}{\epsilon_0} \), which corresponds to option (B) (q / \( 2 \epsilon_{0} \) ).

Step by step solution

01

Understand the problem and Gauss's law

We are given a charge q at the center of the open end of a cylindrical vessel. To find the electric flux through the surface of the vessel, we will use Gauss's law, which states that the total electric flux Φ through a closed surface is given by: Φ = \( \frac{Q_{enclosed}}{\epsilon_0} \) where Q_enclosed is the total charge enclosed by the surface, and ε_0 is the vacuum permittivity. Notice that the vessel is open, so we will have to modify our approach slightly to account for the open end later.
02

Choose a suitable Gaussian surface

To apply Gauss's law, we will need to choose a suitable Gaussian surface. Since the problem involves cylindrical symmetry, it is natural to choose a Gaussian surface in the shape of a cylinder. We will choose the Gaussian cylinder to be concentric with the actual cylindrical vessel and with the same height and radius.
03

Calculate the electric flux through the Gaussian surface

Since we chose a Gaussian cylinder with the same dimensions as the actual cylindrical vessel, the electric flux through the Gaussian surface will be the same as that through the vessel's surface. From Gauss's law, the electric flux Φ through the Gaussian surface is given by: Φ = \( \frac{Q_{enclosed}}{\epsilon_0} \) The charge q at the open end of the vessel is entirely enclosed within the Gaussian cylinder, so Q_enclosed = q. Thus, the electric flux Φ = \( \frac{q}{\epsilon_0} \)
04

Account for the open end of the vessel

We calculated the electric flux through the Gaussian surface, but the actual problem involves an open cylindrical vessel. We must account for the open end to determine the flux through the surface of the vessel. Since the charge q is placed at the center of the open end of the vessel, it is reasonable to assume that the electric flux is uniformly distributed over the closed surface of the Gaussian cylinder. Therefore, the electric flux through the surface of the open cylindrical vessel will be half of the total electric flux calculated above: Φ_vessel = \( \frac{1}{2} \) Φ = \( \frac{1}{2} \) \( \frac{q}{\epsilon_0} \)
05

Match the result with the given options

The electric flux through the surface of the open cylindrical vessel is: Φ_vessel = \( \frac{1}{2} \) \( \frac{q}{\epsilon_0} \) Comparing our result with the given options, we find that our result matches with option (B): (B) (q / \( 2 \epsilon_{0} \) ) So, the correct answer is (B) (q / \( 2 \epsilon_{0} \) ).

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Most popular questions from this chapter

For the system shown in figure, if the resultant force on q is zero, then \(q=\ldots \ldots \ldots\) (A) \(-2 \sqrt{2} \mathrm{Q}\) (B) \(2 \sqrt{2} \mathrm{Q}\) (C) \(2 \sqrt{3} \mathrm{Q}\) (D) \(-3 \sqrt{2} Q\)

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