The Question is from neet-physics electrostatics , and it was created by the author NAGA SAI

Question

A spherical capacitor with inner radius 'a' and outer radius 'b' is filled with a dielectric of permittivity $\epsilon$. A second spherical capacitor with inner radius 'b' and outer radius 'c' is filled with a dielectric of permittivity $2\epsilon$.  If these two capacitors are connected in series, what is the equivalent capacitance of the combination?

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Capacitance of spherical capacitors and series combination of capacitors.
	**Related Topics**: Dielectric constant, equivalent capacitance.
	**Prerequisites**: Formula for capacitance of a spherical capacitor, formula for equivalent capacitance of series capacitors.
## SOLUTION PATHWAY
**Given Information**:
	Capacitor 1: Inner radius 'a', outer radius 'b', permittivity $\epsilon$.
	Capacitor 2: Inner radius 'b', outer radius 'c', permittivity $2\epsilon$.
	Capacitors are connected in series.
**Method & Steps**:
	1. Calculate the capacitance of the first spherical capacitor.
		Reasoning: The capacitance of a spherical capacitor with inner radius $r_1$ and outer radius $r_2$ filled with a dielectric of permittivity $\epsilon$ is given by $C = \frac{4\pi \epsilon r_1 r_2}{r_2 - r_1}$.
		Formula/Rule: $C = \frac{4\pi \epsilon r_1 r_2}{r_2 - r_1}$
	2. Calculate the capacitance of the second spherical capacitor.
		Working: Using the same formula, $C_2 = \frac{4\pi (2\epsilon)bc}{c - b} = \frac{8\pi \epsilon bc}{c - b}$.
		Key Point: The permittivity is $2\epsilon$.
	3. Calculate the equivalent capacitance of the series combination.
		Working: For capacitors in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$. Substituting $C_1$ and $C_2$ we get $\frac{1}{C_{eq}} = \frac{b-a}{4\pi\epsilon ab} + \frac{c-b}{8\pi\epsilon bc} = \frac{2bc(b-a)+ab(c-b)}{8\pi\epsilon abc}$. Therefore, $C_{eq} = \frac{8\pi\epsilon abc}{2bc - 2ac + ac - ab} = \frac{8\pi\epsilon abc}{2b^2 - 2ab + ac - ab} = \frac{8\pi\epsilon abc}{2b^2 - 3ab + ac}$.
		Result: $C_{eq} = \frac{8\pi\epsilon abc}{2b^2 - 3ab + ac}$
		Verification: The units of the expression should be Farads which is consistent with the answer.
 ## OPTION ANALYSIS
[A] ✓: The final derived capacitance matches the formula in option A.
[B] :x: The denominator is incorrect.
[C] :x: The denominator is incorrect.
[D] :x: The denominator is incorrect.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting to invert the sum of reciprocal capacitances when calculating equivalent series capacitance. To avoid this, ensure the final answer is the reciprocal of the sum of reciprocal capacitances.
		2.[Error 2]: Using incorrect formula for spherical capacitor capacitance. To avoid this, remember the correct formula with the difference of radii in the denominator.
	**Quick Tips**:
		1.[Helpful shortcut]: When dealing with series capacitors, it can be useful to combine the fractions before inverting for the final answer.
		2.[Memory aid]: Remember that equivalent series capacitance is always less than the smallest individual capacitance.

NAGA SAI · Answer

The correct Option is "\frac{8\pi\epsilon abc}{2b^2 - 3ab + ac}"

NAGA SAI · Answer

\frac{4\pi\epsilon abc}{2b^2 - ab + ac}

NAGA SAI · Answer

\frac{8\pi\epsilon abc}{b^2 - 2ab + 2ac}

NAGA SAI · Answer

\frac{4\pi\epsilon abc}{b^2 - ab + c}

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