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(D) To find the electric field due to the remaining portion,we use the principle of superposition by considering the sphere as a combination of a full

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$\frac{18}{34}$

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(D) To find the electric field due to the remaining portion,we use the principle of superposition by considering the sphere as a combination of a full sphere of charge density $+ho$ and a smaller sphere of charge density $-ho$.At point $A$ (which is at the center of the carved-out sphere and at a distance $R/2$ from the center of the large sphere):$|\overrightarrow{E}_{A}| = |\overrightarrow{E}_{large} + \overrightarrow{E}_{small}| = |\frac{ho (R/2)}{3\epsilon_0} + 0| = \frac{ho R}{6\epsilon_0}$.At point $B$ (which is at the bottom of the large sphere,distance $R$ from its center and $3R/2$ from the center of the carved-out sphere):$|\overrightarrow{E}_{B}| = |\frac{ho R}{3\epsilon_0} - \frac{ho (R/2)^3}{3\epsilon_0 (3R/2)^2}| = \frac{ho R}{3\epsilon_0} - \frac{ho R^3/8}{3\epsilon_0 (9R^2/4)} = \frac{ho R}{3\epsilon_0} (1 - \frac{1}{18}) = \frac{ho R}{3\epsilon_0} (\frac{17}{18}) = \frac{17ho R}{54\epsilon_0}$.Taking the ratio:$\frac{|\overrightarrow{E}_{A}|}{|\overrightarrow{E}_{B}|} = \frac{ho R / 6\epsilon_0}{17ho R / 54\epsilon_0} = \frac{1}{6} 	imes \frac{54}{17} = \frac{9}{17} = \frac{18}{34}$.

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