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Step-by-Step Solution
Step 1: Understand the Problem
A uniformly charged sphere of radius $R$ with charge density $\rho$ has a smaller sphere of radius $R/2$ carved out from its interior. We need to find the ratio of the magnitudes of the electric field due to the remaining portion of the bigger sphere at two specific points: point A (the center of the big sphere) and point B (a point on the surface of the big sphere).
Step 2: Electric Field Due to a Uniformly Charged Sphere
Complete sphere of radius R:
Inside a uniformly charged sphere, the electric field at any internal point varies linearly with distance from the center. At the center ($r=0$), the electric field is zero, while on the surface ($r=R$) the magnitude of the electric field is
$$
E_{\text{sphere, surface}} = \frac{k\,\rho\,\frac{4}{3}\pi R^{3}}{R^{2}} \;=\; k\,\rho\,\frac{4}{3}\pi R.
$$
Here $k = \frac{1}{4\pi \varepsilon_0}$ for simplicity of notation.
Step 3: Effect of the Removed Sphere
When the smaller sphere of radius $\frac{R}{2}$ is removed, the field at any point is the field due to the full sphere of radius $R$ minus the field that the removed sphere (of radius $\frac{R}{2}$) would have produced if it were still present. We analyze two specific points:
Point A (Center of the original sphere):
The electric field at the center of the complete sphere is $0.$
The removed portion (of radius $R/2$) contributes an electric field at the center of the bigger sphere equivalent to what the center of that smaller sphere would have produced at Point A. However, because Point A lies exactly where the smaller sphere was “touching” the center, the effect is as if we are looking at the field on the surface of the removed sphere (distance $R/2$ from the smaller sphere's center). Mathematically:
$$
E_{\text{removed, at A}} \;=\; \frac{k\,\rho\,\frac{4}{3}\pi \left(\tfrac{R}{2}\right)^{3}}{\left(\tfrac{R}{2}\right)^{2}}
\;=\; k\,\rho\,\frac{4}{3}\pi \left(\frac{R}{2}\right).
$$
Thus,
$$
\left|\overrightarrow{E_A}\right|
= \underbrace{0}_{\text{Full sphere at center}}
\;-\; \underbrace{k\,\rho\,\frac{4}{3}\pi \left(\tfrac{R}{2}\right)}_{\text{removed portion at A}}
= k\,\rho\,\frac{4}{3}\pi \left(\frac{R}{2}\right).
$$
(The negative sign reflects that we remove this contribution; effectively, the net field magnitude is as shown on the right-hand side.)
Point B (Surface of the original sphere):
The electric field at the surface of the complete sphere is
$$
E_{\text{sphere, surface}}
= \frac{k\,\rho\,\frac{4}{3}\pi R^{3}}{R^{2}}
= k\,\rho\,\frac{4}{3}\pi R.
$$
The contribution from the removed sphere (of radius $\tfrac{R}{2}$) at point B is the field due to that smaller sphere at a distance $r = \tfrac{3R}{2}$ (since the center of the removed sphere is $R/2$ away from the big sphere's center, and point B is $R$ away from the big sphere's center, for a total of $R + \tfrac{R}{2} = \tfrac{3R}{2}$ from the center of the smaller sphere). That field is
$$
E_{\text{removed, at B}}
= \frac{k\,\rho\,\frac{4}{3}\pi \left(\tfrac{R}{2}\right)^{3}}{\left(\tfrac{3R}{2}\right)^{2}}
= k\,\rho\,\frac{4}{3}\pi \,\frac{\left(\tfrac{R}{2}\right)^3}{\left(\tfrac{3R}{2}\right)^2}
= k\,\rho\,\frac{4}{3}\pi \,\frac{\tfrac{R^3}{8}}{\tfrac{9R^2}{4}}
= k\,\rho\,\frac{4}{3}\pi \,\frac{R}{18}.
$$
Hence, the net field at point B after removing the smaller sphere becomes
$$
\left|\overrightarrow{E_B}\right|
= \underbrace{k\,\rho\,\frac{4}{3}\pi R}_{\text{full sphere at surface}}
\;-\; \underbrace{k\,\rho\,\frac{4}{3}\pi \,\frac{R}{18}}_{\text{removed portion at B}}
= k\,\rho\,\frac{4}{3}\pi \left( R - \frac{R}{18} \right)
= k\,\rho\,\frac{4}{3}\pi \left(\frac{17R}{18}\right).
$$
Step 4: Compute the Required Ratio
The ratio of the magnitudes of the electric fields at points A and B is
$$
\frac{\left|\overrightarrow{E_A}\right|}{\left|\overrightarrow{E_B}\right|}
= \frac{k\,\rho\,\frac{4}{3}\pi \left(\frac{R}{2}\right)}{k\,\rho\,\frac{4}{3}\pi \left(\frac{17R}{18}\right)}
= \frac{\frac{R}{2}}{\frac{17R}{18}}
= \frac{18}{34}
= \frac{9}{17}.
$$
Step 5: Final Answer
The required ratio is
$$
\frac{\left|\overrightarrow{E_A}\right|}{\left|\overrightarrow{E_B}\right|}
= \frac{18}{34}.
$$
