Use the following figure to answer the next problem: -A sphere of radius 8.0 cm carries a uniform volume charge density $\rho$ = 500 nC/m³.What is the electric field at r = 20 cm? A)0.24 kN/C B)0.12 kN/C C)4.8 N/C D)15 N/C E)5.4 N/C

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The electric field inside a uniformly charged sphere at a distance $r$ from the center is given by $E = \frac{1}{4\pi\epsilon_0} \frac{\rho r}{R^3}$, where $\rho$ is the volume charge density, $R$ is the radius of the sphere, and $r$ is the distance from the center of the sphere to the point where the electric field is being calculated. However, for $r > R$, the electric field is calculated as if all the charge were concentrated at the center of the sphere, using the formula $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$, where $Q$ is the total charge and $r$ is the distance from the center. Given $\rho = 500 \times 10^{-9} \, \text{C/m}^3$ and $R = 8 \, \text{cm} = 0.08 \, \text{m}$, the total charge $Q = \rho \times \frac{4}{3}\pi R^3$. For $r = 20 \, \text{cm} = 0.2 \, \text{m}$, the electric field $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$ results in $E = 0.24 \, \text{kN/C}$ after substituting the values and calculating.

Use the Following Figure to Answer the Next Problem ρ\rhoρ = 500 NC/m3

Use the Following Figure to Answer the Next Problem $\rho$ = 500 NC/m³