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SI123 Group Title

Three concentric spherical shells have radii a, b and c(a<b<c) and have surface charge densities σ, -σ and σ respectively. If Va, Vb and Vc denote the potentials of the three shells, then for c=a+b we have (1)Vc=Va≠Vb (2)Vc=Vb≠Va (3)Vc≠Vb≠Va (4)Vc=Vb=Va

  • one year ago
  • one year ago

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  1. Aqua666 Group Title
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    I think it's 3 but I can't seem to explain why.. It's a matter of reasoning going from the basis that the potentials are dependent on both radius and surface charge. Looking at the information given, there seems to be no way that any of the potentials could ever be equal to one of the others without disregarding one of the equations.

    • one year ago
  2. SI123 Group Title
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    The answer in book is 1 but it can be a misprint as well though a small solution is also given in the book. Still it is beyond my understanding. I am giving the solution as given in book........\[Va=\frac{ 1 }{ 4\pi \epsilon _{o} }\frac{ \sigma4\pi a ^{2} }{ a }-\frac{ 1 }{ 4\pi \epsilon _{o} }\frac{ \sigma4\pi b ^{2} }{ b }+\frac{ 1 }{ 4\pi \epsilon _{o} }\frac{ \sigma4\pi c ^{2} }{ c }\] \[V _{b}=\frac{ 1 }{ 4\pi \epsilon _{o} }\frac{ \sigma4\pi a ^{2} }{ a }-\frac{ 1 }{ 4\pi \epsilon _{o} }\frac{ \sigma4\pi b ^{2} }{ b }+-\frac{ 1 }{ 4\pi \epsilon _{o} }\frac{ \sigma4\pi c ^{2} }{ c }\] \[V _{c}=\frac{ 1 }{ 4\pi \epsilon _{o} }\frac{ \sigma4\pi a ^{2} }{ a }-\frac{ 1 }{ 4\pi \epsilon _{o} }\frac{ \sigma4\pi b ^{2} }{ b }+\frac{ 1 }{ 4\pi \epsilon _{o} }\frac{ \sigma4\pi c ^{2} }{ c }\] Hence Va=Vb≠Vc

    • one year ago
  3. Shadowys Group Title
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    for c=a+b \(V_a=\frac{1}{4\pi \epsilon_0}( \frac{4\pi a^2 \sigma }{a}-\frac{4\pi b^2 \sigma }{b}+\frac{4\pi c^2 \sigma }{c})=\frac{\sigma}{\epsilon_0}(a-b+c)=\frac{\sigma}{\epsilon_0}(2a)\) \(V_b=\frac{1}{4\pi \epsilon_0}( \frac{4\pi a^2 \sigma }{b}-\frac{4\pi b^2 \sigma }{b}+\frac{4\pi c^2 \sigma }{c})=\frac{\sigma}{\epsilon_0}(\frac{a^2}{b}-b+c)=\frac{\sigma}{\epsilon_0}(\frac{a^2 +ab}{b})\) \(V_c=\frac{1}{4\pi \epsilon_0}( \frac{4\pi a^2 \sigma }{c}-\frac{4\pi b^2 \sigma }{c}+\frac{4\pi c^2 \sigma }{c})=\frac{\sigma}{\epsilon_0}(\frac{a^2-b^2}{a+b}+a+b)=\frac{\sigma}{\epsilon_0}(2a)\) thus answer is (1) do you need further help?

    • one year ago
  4. Shadowys Group Title
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    note that the potential outside the charged sphere is \(V=\frac{r_0^2 \sigma}{\epsilon_0 R}\) but inside the sphere it's\(V=\frac{r_0 \sigma}{\epsilon_0 }\)

    • one year ago
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