Consider a solid sphere with radius $R$ with a charge uniformly spread throughout its volume. The magnitude of the sphere's electric field is $\left|\vec{E}_{1}\right|$ at a point $P_{1}$ on the sphere's surface and is $\left|\vec{E}_{2}\right|$ at a point $P_{2}$ at a distance $\frac{1}{2} R$ from the sphere's center. In the relationship $\left|\vec{E}_{1}\right|=b\left|\vec{E}_{2}\right|$, what is the value of $b$ ?
A) 0
B) 1
C) 2
D) 4
E) 8
F) Other

Question

Quizplus · Accepted Answer

The electric field inside a uniformly charged sphere increases linearly with distance from the center, reaching its maximum at the surface. At a distance $r$ from the center, the magnitude of the electric field is given by $\left|\vec{E}\right| = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^3} r$ for $r \leq R$, where $Q$ is the total charge, $R$ is the radius of the sphere, and $\epsilon_0$ is the vacuum permittivity. At the surface ($r = R$), this becomes $\left|\vec{E}_1\right| = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}$. At a distance $\frac{1}{2}R$ from the center, the electric field is $\left|\vec{E}_2\right| = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^3} \left(\frac{1}{2}R\right) = \frac{1}{2} \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}$. Therefore, $\left|\vec{E}_1\right| = 2\left|\vec{E}_2\right|$, making $b = 2$.

Consider a Solid Sphere with Radius RRR With a Charge Uniformly Spread Throughout Its Volume

Consider a Solid Sphere with Radius $R$ With a Charge Uniformly Spread Throughout Its Volume