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Question

Two concentric conducting spherical shells of radii $a$ and $b$ ($b > a$) carry charges $+Q$ and $-Q$ respectively. The electric field at a distance $r$ from the center, where $a < r < b$, is:

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
 **Primary Concept**: Gauss's Law for electricity
 **Related Topics**: Electric field of spherical charge distributions, conductors
 **Prerequisites**: Understanding of Gauss's law, electric flux, and properties of conductors.
## SOLUTION PATHWAY
**Given Information**:
  * Two concentric conducting spherical shells with radii $a$ and $b$ ($b > a$)
  * Inner shell has charge $+Q$
  * Outer shell has charge $-Q$
  * We need to find the electric field at a distance $r$ from the center, where $a < r < b$
**Method & Steps**:
 1. Apply Gauss's Law:  Gauss's law states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space ($\epsilon_0$).  We'll use a Gaussian surface in the form of a sphere with radius $r$, where $a < r < b$.
  Reasoning: Since we are considering the region between the shells, the Gaussian surface encloses only the charge $+Q$ on the inner shell.
  Formula/Rule: $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$
 2. Calculate the electric flux: The electric field is radial and has the same magnitude at every point on the Gaussian surface. Thus, the dot product simplifies, and the integral becomes:
  Working: $E \oint dA = \frac{Q}{\epsilon_0}$, where $\oint dA = 4\pi r^2$ (surface area of the sphere)
  Key Point: The enclosed charge ($Q_{enc}$) is $+Q$ because the Gaussian surface is between the shells.
 3. Solve for the electric field:
  Result: $E(4\pi r^2) = \frac{Q}{\epsilon_0} \Rightarrow E = \frac{Q}{4\pi\epsilon_0 r^2}$ radially outward
  Verification: The formula shows that the field strength decreases with the square of the distance, which is consistent with the inverse square law for point charges. The direction is radially outward because the enclosed charge is positive.
 ## OPTION ANALYSIS
[A] :x: Zero: This is incorrect because a charge is enclosed within the Gaussian surface.
[B] :x: $\frac{Q}{4\pi\epsilon_0 a^2}$ radially outward: This would be the field at the surface of the inner shell, not at an arbitrary distance $r$ between the shells.
[C] ✓: $\frac{Q}{4\pi\epsilon_0 r^2}$ radially outward: This is the correct answer derived from Gauss's law.
[D] :x: $\frac{Q}{4\pi\epsilon_0 b^2}$ radially inward: This would be related to the field at the surface of the outer shell, but the direction is incorrect, and it doesn't consider the field at a general point between the shells.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly applying Gauss's Law: Forgetting to consider only the enclosed charge or miscalculating the flux.
  2.Confusing the field inside a conductor with the field between conductors:  The electric field inside a conductor is zero in electrostatic equilibrium, but this is not the case in the space between the shells.
 **Quick Tips**:
  1.Always draw a Gaussian surface and clearly identify the enclosed charge.
  2.Remember that the electric field inside a conductor is zero in electrostatic equilibrium.

NAGA SAI · Answer

Zero

NAGA SAI · Answer

$\frac{Q}{4\pi\epsilon_0 a^2}$ radially outward

NAGA SAI · Answer

The correct Option is "$\frac{Q}{4\pi\epsilon_0 r^2}$ radially outward"

NAGA SAI · Answer

$\frac{Q}{4\pi\epsilon_0 b^2}$ radially inward

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