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Chapter 11: Problem 1564

A sphere of radius \(R\) has a uniform distribution of electric charge in its volume. At a distance \(\mathrm{x}\) from its centre, (for \(\mathrm{x}<\mathrm{R})\), the electric field is directly proportional to ...... (A) \(\mathrm{x}\) (B) \(\mathrm{x}^{-1}\) (C) \(x^{-2}\) (D) \(\mathrm{x}^{2}\)

Short Answer

Expert verified
(A) \(\mathrm{x}\)

Step by step solution

01

Understanding Gauss's Law

Gauss's Law states that the electric flux through a closed surface is proportional to the total charge enclosed by the surface. It is given by the equation: \[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}\] where \(\oint \vec{E} \cdot d\vec{A}\) is the electric flux, \(Q_{enc}\) is the charge enclosed by the surface, and \(\epsilon_0\) is the vacuum permittivity.
02

Use a spherical Gaussian surface

To apply Gauss's Law to the charged sphere, we will choose a Gaussian surface in the shape of a sphere with radius x. Its surface area is given by: \[A = 4\pi x^2\] This surface is symmetric about the center of the charged sphere, allowing us to ignore any angular dependence in our calculation.
03

Calculate the enclosed charge

Since the electric charge is uniformly distributed in the sphere, we can calculate the charge density \(\rho\): \[\rho = \frac{Q_{total}}{\frac{4}{3}\pi R^3}\] Now, we will determine the enclosed charge by the Gaussian surface. We know that: \[Q_{enc} = \rho \cdot V_{enc}\] Since the enclosed volume is given by the volume of the Gaussian surface: \[V_{enc} = \frac{4}{3}\pi x^3\] The enclosed charge becomes: \[Q_{enc} = \rho \cdot V_{enc} = \frac{Q_{total}}{\frac{4}{3}\pi R^3} \cdot \frac{4}{3}\pi x^3\]
04

Apply Gauss's Law and find the electric field

Now, we will use Gauss's Law to find the electric field \(\vec{E}\) inside the sphere. As the Gaussian surface is symmetric, the electric field \(\vec{E}\) will be constant at any point on its surface (at a distance x from the center). Therefore, we can write Gauss's Law as: \[\vec{E} \cdot A = \frac{Q_{enc}}{\epsilon_0}\] Substituting the values of A and \(Q_{enc}\), we get: \[\vec{E} \cdot 4\pi x^2 = \frac{Q_{total}}{\frac{4}{3}\pi R^3} \cdot \frac{4}{3}\pi x^3 \cdot \frac{1}{\epsilon_0}\] Solving for \(\vec{E}\), we obtain the following relationship between electric field and x: \[\vec{E} = \frac{Q_{total}}{4\pi \epsilon_0 R^3} \cdot x\]
05

Analyze the relationship between electric field and x

Comparing the obtained relationship with the given options, we can see that the electric field is directly proportional to x, as it has a linear dependency on x: \[\vec{E} \propto x\] Thus, the correct answer is: (A) \(\mathrm{x}\)

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Most popular questions from this chapter

Two thin wire rings each having a radius \(R\) are placed at a distance \(d\) apart with their axes coinciding. The charges on the two rings are \(+q\) and \(-q\). The potential difference between the centers of the two rings is $\ldots .$ (A) 0 (B) $\left.\left[\mathrm{q} /\left(2 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{(}^{2}+\mathrm{d}^{2}\right)\right\\}\right]$ (C) $\left[\mathrm{q} /\left(4 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{\left. \left.\left(\mathrm{R}^{2}+\mathrm{d}^{2}\right)\right\\}\right]}\right.\right.$ (D) \(\left[(q R) /\left(4 \pi \epsilon_{0} d^{2}\right)\right]\)

4 Points charges each \(+q\) is placed on the circumference of a circle of diameter \(2 \mathrm{~d}\) in such a way that they form a square. The potential at the centre is \(\ldots \ldots .\) (A) 0 (B) \((4 \mathrm{kd} / \mathrm{q})\) (C) \((\mathrm{kd} / 4 \mathrm{q})\) (D) \((4 \mathrm{kq} / \mathrm{d})\)

A thin spherical shell of radius \(R\) has charge \(Q\) spread uniformly over its surface. Which of the following graphs, figure most closely represents the electric field \(\mathrm{E}\) (r) produced by the shell in the range $0 \leq \mathrm{r}<\infty\(, where \)\mathrm{r}$ is the distance from the centre of the shel1.

Three charges, each of value \(Q\), are placed at the vertex of an equilateral triangle. A fourth charge \(q\) is placed at the centre of the triangle. If the charges remains stationery then, \(q=\ldots \ldots \ldots\) (A) \((\mathrm{Q} / \sqrt{2})\) (B) \(-(\mathrm{Q} / \sqrt{3})\) (C) \(-(Q / \sqrt{2})\) (D) \((\mathrm{Q} / \sqrt{3})\)

Two identical metal plates are given positive charges \(\mathrm{Q}_{1}\) and \(\mathrm{Q}_{2}\left(<\mathrm{Q}_{1}\right)\) respectively. If they are now brought close to gather to form a parallel plate capacitor with capacitance \(\mathrm{c}\), the potential difference between them is (A) \(\left[\left(Q_{1}+Q_{2}\right) /(2 c)\right]\) (B) \(\left[\left(\mathrm{Q}_{1}+\mathrm{Q}_{2}\right) / \mathrm{c}\right]\) (C) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) /(2 \mathrm{c})\right]\) (D) \(\left[\left(\mathrm{Q}_{1}-\mathrm{Q}_{2}\right) / \mathrm{c}\right]\)
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