A non-conducting solid sphere has radius R an | vedclass

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(C) Let $Q_0$ be the charge of the solid sphere before the cavity is created. The electric field at any point inside a uniformly charged non-conductin

Vedclass · Accepted Answer

$0.51$

Vedclass · Answer

(C) Let $Q_0$ be the charge of the solid sphere before the cavity is created. The electric field at any point inside a uniformly charged non-conducting sphere is given by $\vec{E} = \frac{ho \vec{r}}{3 \epsilon_0} = \frac{Q_0 \vec{r}}{4 \pi \epsilon_0 R^3}$.By the principle of superposition,the electric field at the center of the cavity $(O')$ is the field due to the original sphere minus the field that would be produced by the removed spherical part.$E = E_{Q_0} - E_{	ext{cavity}} = \frac{Q_0}{4 \pi \epsilon_0 R^3} \left( \frac{R}{2} ight) - 0 = \frac{Q_0}{8 \pi \epsilon_0 R^2} = \frac{Q_0}{4 \pi \epsilon_0 R^2} \left( \frac{1}{2} ight)$.Since the charge density $ho$ is uniform,the charge is proportional to the volume: $\frac{Q}{Q_0} = \frac{V_{	ext{sphere}} - V_{	ext{cavity}}}{V_{	ext{sphere}}} = \frac{\frac{4}{3} \pi R^3 - \frac{4}{3} \pi (R/4)^3}{\frac{4}{3} \pi R^3} = 1 - \frac{1}{64} = \frac{63}{64}$.Thus,$Q_0 = Q \left( \frac{64}{63} ight)$.Substituting $Q_0$ into the expression for $E$:$E = \frac{Q (64/63)}{4 \pi \epsilon_0 R^2} \left( \frac{1}{2} ight) = \frac{32}{63} \left( \frac{Q}{4 \pi \epsilon_0 R^2} ight) \approx 0.5079 \left( \frac{Q}{4 \pi \epsilon_0 R^2} ight)$.Comparing this with $E = K \left( \frac{Q}{4 \pi \epsilon_0 R^2} ight)$,we get $K \approx 0.51$.

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