Three concentric spherical metallic shells X, | vedclass

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(D) The charges on the shells are:$q_x = \sigma(4\pi a^2)$$q_y = -\sigma(4\pi b^2)$$q_z = \sigma(4\pi c^2)$Since the potential of shell $X$ is equal t

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(D) The charges on the shells are:$q_x = \sigma(4\pi a^2)$$q_y = -\sigma(4\pi b^2)$$q_z = \sigma(4\pi c^2)$Since the potential of shell $X$ is equal to the potential of shell $Z$ $(V_x = V_z)$:$V_x = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_x}{a} + \frac{q_y}{b} + \frac{q_z}{c} \right)$$V_z = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_x}{c} + \frac{q_y}{c} + \frac{q_z}{c} \right)$Equating $V_x$ and $V_z$:$\frac{\sigma 4\pi a^2}{a} - \frac{\sigma 4\pi b^2}{b} + \frac{\sigma 4\pi c^2}{c} = \frac{\sigma 4\pi a^2}{c} - \frac{\sigma 4\pi b^2}{c} + \frac{\sigma 4\pi c^2}{c}$Simplifying the equation:$a - b + \frac{c^2}{c} = \frac{a^2 - b^2 + c^2}{c}$$a - b + c = \frac{a^2 - b^2 + c^2}{c}$$c(a - b + c) = a^2 - b^2 + c^2$$c(a - b) + c^2 = a^2 - b^2 + c^2$$c(a - b) = a^2 - b^2$$c(a - b) = (a - b)(a + b)$$c = a + b$Given $a = 2\,cm$ and $b = 3\,cm$:$c = 2 + 3 = 5\,cm$.

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