A solid sphere of radius R_1 and volume charg | vedclass

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Question

The total positive charge on the inner solid sphere

$\int_{0}^{R_{1}} 4 \pi r^{2} d r \cdot \frac{\rho_{0}}{r}=+4 \pi \frac{R_{1}^{2}}{2} \rho_{0}$

Vedclass · Accepted Answer

$\sqrt {\frac{{{\rho _0}}}{{2\sigma }}} $

Vedclass · Answer

The total positive charge on the inner solid sphere

$\int_{0}^{R_{1}} 4 \pi r^{2} d r \cdot \frac{ho_{0}}{r}=+4 \pi \frac{R_{1}^{2}}{2} ho_{0}$

The total negative charge on the outer shell $=-4 \pi \mathrm{R}_{2}^{2} \sigma$ but both are equal in magnitude as the sum is zero

$\frac{\mathrm{R}_{2}^{2}}{\mathrm{R}_{1}^{2}}=\frac{ho_{0}}{2 \sigma} \quad 	herefore \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\sqrt{\frac{ho_{0}}{2 \sigma}}$

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