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 (a) VC = VB ≠ VA (b) VC ≠ VB ≠ VA (c) VC = VB = VA (d) VC = VA ≠ VB

Answer: (d). Charges: q ∝ ±σ r². Potential of a shell = (own + inner charges)/own radius + outer charges/their radii. VA = (σ)/(ε₀)(a - b + c), VB = (σ)/(ε₀)(a²/b - b + c), VC = (σ)/(ε₀)·(a² - b² + c²)/(c) With c = a + b: VA = (2aσ)/(ε₀), and a² - b² + c² = 2a(a+b) = 2ac ⇒ VC = (2aσ)/(ε₀). VB = (σ)/(ε₀)(a²/b + a) ≠ (2aσ)/(ε₀) (unless a = b). So the answer is (d) VC = VA ≠ VB.

Three concentric spherical shells have radii a, b and c (a < b < c)… | NEET 2009 (AIPMT)

Q120NEET 2009 · AIPMTMedium

Electric Potential & Equipotentials

Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities $\sigma, -\sigma$ and $\sigma$ respectively. If VA, VB and VC denote the potentials of the three shells, then, for c = a + b, we have

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Answer: (d)

Charges: $q \propto \pm\sigma r^2$ . Potential of a shell = (own + inner charges)/own radius + outer charges/their radii.
$V_A = \frac{\sigma}{\varepsilon_0}(a - b + c)$ , $V_B = \frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{b} - b + c\right)$ , $V_C = \frac{\sigma}{\varepsilon_0}\cdot\frac{a^2 - b^2 + c^2}{c}$
With $c = a + b$ : $V_A = \frac{2a\sigma}{\varepsilon_0}$ , and $a^2 - b^2 + c^2 = 2a(a+b) = 2ac \Rightarrow V_C = \frac{2a\sigma}{\varepsilon_0}$ .
$V_B = \frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{b} + a\right) \neq \frac{2a\sigma}{\varepsilon_0}$ (unless $a = b$ ).
So the answer is (d) $V_C = V_A \neq V_B$ .

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