Three concentric spherical metallic shells A- B and C of radii a- b and c a - b- c have charge densities - - and - respectively- If the shells A and C are at the same potential then the relation between a- b and c is -

The correct option is C a + b = c #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; ...

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Standard XII

Chemistry

Depression in Freezing Point

Three concent...

Question

Three concentric spherical metallic shells A, B and C of radii $a, b$ and $c (a < b < c)$ have charge densities $σ, - σ$ and $σ$ respectively. If the shells A and C are at the same potential then the relation between a, b and c is :

a + b + c = 0

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a + c = b

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a + b = c

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a = b + c

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Solution

The correct option is C $a + b = c$

$V_{A} = \frac{σ 4 π a^{2}}{4 π ϵ_{0} a} + \frac{(- σ) 4 π b^{2}}{4 π ϵ_{0} b} + \frac{σ 4 π c^{2}}{4 π ϵ_{0} c}$

$= \frac{σ}{ϵ_{0}} (a - b + c)$

Similarly, $V_{C} = \frac{σ}{ϵ_{0}} (\frac{a^{2}}{c} - \frac{b^{2}}{c} + c)$

But $V_{A} = V_{C}$

$a - b + c = \frac{a^{2} - b^{2}}{c} + c$

Therefore, $a + b = c$

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Three concentric spherical metallic shells A, B and C of radii a,b and c (a<b<c) have charge densities σ,−σ and σ respectively. If the shells A and C are at the same potential then the relation between a, b and c is :

The correct option is C a+b=c VA=σ4πa24πϵ0a+(−σ)4πb24πϵ0b+σ4πc24πϵ0c =σϵ0(a−b+c) Similarly, VC=σϵ0(a2c−b2c+c) But VA=VC a−b+c=a2−b2c+c Therefore, a+b=c

Three concentric spherical metallic shells A, B and C of radii $a, b$ and $c (a < b < c)$ have charge densities $σ, - σ$ and $σ$ respectively. If the shells A and C are at the same potential then the relation between a, b and c is :