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(B) Let $\rho$ be the uniform charge density of the solid sphere. The electric field at any point $P$ inside the cavity can be calculated by the princ

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Non-zero and uniform

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(B) Let $ho$ be the uniform charge density of the solid sphere. The electric field at any point $P$ inside the cavity can be calculated by the principle of superposition,treating the cavity as a sphere with charge density $-ho$ superimposed on the original solid sphere with charge density $+ho$.$\vec{E}_{P} = \vec{E}_{	ext{sphere}} + \vec{E}_{	ext{cavity}}$Using the formula for the electric field inside a uniformly charged sphere,$\vec{E} = \frac{ho \vec{r}}{3 \varepsilon_0}$,we have:$\vec{E}_{P} = \frac{ho \vec{r}_1}{3 \varepsilon_0} + \frac{(-ho) \vec{r}_2}{3 \varepsilon_0} = \frac{ho}{3 \varepsilon_0} (\vec{r}_1 - \vec{r}_2)$From the geometry of the figure,by the triangle law of vector addition,$\vec{r}_1 = \vec{C}_1\vec{C}_2 + \vec{r}_2$,which implies $\vec{r}_1 - \vec{r}_2 = \vec{C}_1\vec{C}_2$.Therefore,$\vec{E}_{P} = \frac{ho}{3 \varepsilon_0} \vec{C}_1\vec{C}_2$.Since $ho$,$\varepsilon_0$,and the vector $\vec{C}_1\vec{C}_2$ (the displacement between the centers of the two spheres) are constants,the electric field $\vec{E}_{P}$ inside the cavity is non-zero and uniform.

$A$ spherical portion has been removed from a solid sphere having a charge distributed uniformly as shown in the figure. The electric field inside the emptied space is $:-$

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