A solid sphere of radius R_1 and volume charg | vedclass

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(C) The total positive charge $Q_{in}$ on the inner solid sphere is calculated by integrating the volume charge density over the volume of the sphere:

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$\sqrt{\frac{\rho_0}{2\sigma}}$

Vedclass · Answer

(C) The total positive charge $Q_{in}$ on the inner solid sphere is calculated by integrating the volume charge density over the volume of the sphere:$Q_{in} = \int_{0}^{R_1} \rho(r) \cdot 4\pi r^2 dr = \int_{0}^{R_1} \frac{\rho_0}{r} \cdot 4\pi r^2 dr = 4\pi \rho_0 \int_{0}^{R_1} r dr = 4\pi \rho_0 \left[ \frac{r^2}{2} \right]_0^{R_1} = 2\pi \rho_0 R_1^2$.The total negative charge $Q_{out}$ on the outer hollow sphere is given by the surface charge density multiplied by its surface area:$Q_{out} = -\sigma \cdot 4\pi R_2^2 = -4\pi \sigma R_2^2$.Given that the total charge in the system is zero, we have $Q_{in} + Q_{out} = 0$, which implies $Q_{in} = |Q_{out}|$.$2\pi \rho_0 R_1^2 = 4\pi \sigma R_2^2$.Rearranging for the ratio $\frac{R_2^2}{R_1^2}$:$\frac{R_2^2}{R_1^2} = \frac{2\pi \rho_0}{4\pi \sigma} = \frac{\rho_0}{2\sigma}$.Taking the square root of both sides, we get:$\frac{R_2}{R_1} = \sqrt{\frac{\rho_0}{2\sigma}}$.

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