Consider a metal sphere enclosed concentrically within a spherical shell. The inner sphere of radius a carries charge Q. The outer shell of radius 2a also has charge Q. The variation of the magnitude E of the electric field as a function of distance r from the center O is
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For concentric spherical shells, remember these rules:
1. E = 0 inside any conducting material.
2. Use Gauss's law with a spherical surface for regions in between or outside.
3. Discontinuities in E occur at surfaces with surface charge. The jump is given by \(\Delta E = \sigma/\epsilon_0\). A positive surface charge causes an upward jump in the E vs r graph.
Step 1: Understanding the Concept: The problem asks for the electric field \(E\) as a function of distance \(r\) for a system of two concentric spherical conductors. The inner sphere has radius \(a\) and charge \(Q\). The outer spherical shell has radius \(2a\) and also has a charge \(Q\). We need to find \(E(r)\) in three regions: \(r<a\), \(a<r<2a\), and \(r>2a\). The key tool is Gauss's Law for spherically symmetric charge distributions. Step 2: Key Formula or Approach: Gauss's Law for a spherical Gaussian surface of radius \(r\) concentric with the charge distribution is: \[ \oint \mathbf{E} \cdot d\mathbf{A} = E(r) \cdot (4\pi r^2) = \frac{Q_{enc}}{\epsilon_0} \] \[ \implies E(r) = \frac{Q_{enc}}{4\pi\epsilon_0 r^2} \] We need to determine the enclosed charge \(Q_{enc}\) for each region. Also, remember that the electric field inside a conductor in electrostatic equilibrium is zero. Step 3: Detailed Explanation: 1. Region 1: \(r<a\) (Inside the inner metal sphere) Since the inner sphere is a conductor, the electric field inside it must be zero in electrostatic equilibrium. \[ E = 0 \quad \text{for } r<a \] 2.
Region 2: \(a<r<2a\) (Between the inner sphere and the outer shell) Draw a Gaussian surface of radius \(r\) such that \(a<r<2a\). The charge enclosed, \(Q_{enc}\), is the charge on the inner sphere, which is \(Q\). \[ E(r) = \frac{Q}{4\pi\epsilon_0 r^2} \quad \text{for } a<r<2a \] This field is non-zero and decreases as \(1/r^2\). At \(r = a\), the field just outside the surface is \(E(a^+) = \frac{Q}{4\pi\epsilon_0 a^2}\). This shows a discontinuity, jumping from 0 to a finite value. 3. Inside the material of the outer shell (radius 2a) The outer shell is also a conductor, so the electric field within its material must be zero. To achieve this, the charge \(Q\) from the inner sphere must induce a charge of \(-Q\) on the inner surface of the outer shell. The outer shell itself has a total charge of \(Q\). So, the charge on its outer surface will be \(Q - (-Q) = 2Q\). 4.
Region 3: \(r>2a\) (Outside the outer shell) Draw a Gaussian surface of radius \(r\) such that \(r>2a\). The total charge enclosed, \(Q_{enc}\), is the sum of the charge on the inner sphere and the charge on the outer shell. \[ Q_{enc} = Q_{\text{inner}} + Q_{\text{outer}} = Q + Q = 2Q \] \[ E(r) = \frac{2Q}{4\pi\epsilon_0 r^2} \quad \text{for } r>2a \] This field also decreases as \(1/r^2\). Let's check the field values at the boundary \(r = 2a\). Just inside the outer shell (\(r \to 2a^-\)), \(E(2a^-) = \frac{Q}{4\pi\epsilon_0 (2a)^2} = \frac{Q}{16\pi\epsilon_0 a^2}\). Just outside the outer shell (\(r \to 2a^+\)), \(E(2a^+) = \frac{2Q}{4\pi\epsilon_0 (2a)^2} = \frac{2Q}{16\pi\epsilon_0 a^2} = \frac{Q}{8\pi\epsilon_0 a^2}\). The field jumps up at \(r=2a\). Specifically, \(E(2a^+) = 2 E(2a^-)\).
Summary of E(r) behavior:
\(0 \le r<a\): \(E = 0\) \(r = a\): Discontinuity, jumps from 0 to \(\frac{Q}{4\pi\epsilon_0 a^2}\) \(a<r<2a\): \(E \propto 1/r^2\), decreases from \(\frac{Q}{4\pi\epsilon_0 a^2}\) to \(\frac{Q}{16\pi\epsilon_0 a^2}\) \(r = 2a\): Discontinuity, jumps from \(\frac{Q}{16\pi\epsilon_0 a^2}\) to \(\frac{2Q}{16\pi\epsilon_0 a^2}\) \(r>2a\): \(E \propto 1/r^2\), decreases from \(\frac{2Q}{16\pi\epsilon_0 a^2}\) Matching with graphs:
All graphs correctly show \(E=0\) for \(r All graphs show a \(1/r^2\) decay for \(a<r<2a\). We need to check the behavior at \(r=2a\). The field must jump upwards. Graph (A): Shows an upward jump at \(r=2a\). Graph (B): Shows a downward jump at \(r=2a\). (Incorrect) Graph (C): Shows a continuous field at \(r=2a\). (Incorrect) Graph (D): Shows a downward jump at \(r=2a\). (Incorrect) Only Graph (A) correctly shows the upward jump in the electric field at \(r=2a\), which is due to the positive charge residing on the outer surface of the shell. Step 4: Final Answer: The electric field is zero inside the inner sphere, varies as \(1/r^2\) between the spheres, and varies as \(2/r^2\) outside the outer shell. There are upward discontinuities at \(r=a\) and \(r=2a\). Graph (A) correctly depicts this behavior.
