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Chapter 11: Problem 1636

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.

Short Answer

Expert verified
The electric field E(r) for a thin spherical shell of radius R with charge Q uniformly spread over its surface is given by: \[E(r) = \begin{cases} 0 & \text{for } r < R \\ \frac{Q}{4 \pi \epsilon_0 r^2} & \text{for } r \geq R \end{cases}\] The graph representing this function should show E(r) as zero for \(r<R\) and follow an inverse square relationship (\(\frac{1}{r^2}\)) as \(r\geq R\).

Step by step solution

01

Gauss's Law states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity constant, which can be mathematically written as: \[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}\] To apply Gauss's Law, we need to choose an appropriate Gaussian surface. Since we are dealing with a sphere, the Gaussian surface will be a sphere concentric with the given thin shell and with radius r. #Step 2: Determine the enclosed charge inside the Gaussian surface#

We can determine the enclosed charge based on the distance r from the center of the shell: - If r<R: The Gaussian surface lies completely inside the shell, and therefore, there is no charge enclosed. \(Q_{enclosed}=0\) - If r≥R: The Gaussian surface entirely encloses the shell, so the enclosed charge is equal to the total charge on the shell, Q. #Step 3: Calculate the electric field based on the enclosed charge#
02

Now, we can use Gauss's Law to calculate the electric field: - If r

Based on our calculations in Steps 3 and 4, we can write the final electric field function as: \[E(r) = \begin{cases} 0 & \text{for } r < R \\ \frac{Q}{4 \pi \epsilon_0 r^2} & \text{for } r \geq R \end{cases}\] Now we can compare this function to the given graphs to choose the most representative one. The graph should show E(r) as zero while r<R and follow an inverse square relationship (1/r^2) as r≥R.

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Most popular questions from this chapter

Four equal charges \(\mathrm{Q}\) are placed at the four corners of a square of each side is ' \(\mathrm{a}\) '. Work done in removing a charge \- Q from its centre to infinity is ....... (A) 0 (B) $\left[\left(\sqrt{2} \mathrm{Q}^{2}\right) /\left(\pi \epsilon_{0} \mathrm{a}\right)\right]$ (C) $\left[\left(\sqrt{2} Q^{2}\right) /\left(4 \pi \epsilon_{0} a\right)\right]$ (D) \(\left[\mathrm{Q}^{2} /\left(2 \pi \epsilon_{0} \mathrm{a}\right)\right]\)

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 parallel plate capacitor with air between the plates has a capacitance of $9 \mathrm{pF}\(. The separation between its plates is \)\mathrm{d}$. The space between the plates is now filled with two dielectrics. One of the dielectric constant \(\mathrm{K}_{1}=3\) and thickness \(\mathrm{d} / 3\) while the other one has dielectric constant \(\mathrm{K}_{2}=6\) and thickness \(2 \mathrm{~d} / 3\). Capacitance of the capacitor is now (A) \(1.8 \mathrm{pF}\) (B) \(20.25 \mathrm{pF}\) (C) \(40.5 \mathrm{pF}\) (D) \(45 \mathrm{pF}\)

A long string with a charge of \(\lambda\) per unit length passes through an imaginary cube of edge \(\ell\). The maximum possible flux of the electric field through the cube will be ....... (A) \(\sqrt{3}\left(\lambda \ell / \in_{0}\right)\) (B) \(\left(\lambda \ell / \in_{0}\right)\) (C) \(\sqrt{2}\left(\lambda \ell / \in_{0}\right)\) (D) \(\left[\left(6 \lambda \ell^{2}\right) / \epsilon_{0}\right]\)

Three identical spheres each having a charge \(\mathrm{q}\) and radius \(R\), are kept in such a way that each touches the other two spheres. The magnitude of the electric force on any sphere due to other two is \(\ldots \ldots \ldots\) (A) $(\mathrm{R} / 2)\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{5} / 4)(\mathrm{q} / \mathrm{R})^{2}$ (B) $\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{2} / 3)(\mathrm{q} / \mathrm{R})^{2}$ (C) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right](\sqrt{3} / 4)(\mathrm{q} / \mathrm{R})^{2}$ (D) $-\left[1 /\left(8 \pi \epsilon_{0}\right)\right](\sqrt{3} / 2)(\mathrm{q} / \mathrm{R})^{2}$
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