A charge of 5.0 pC is distributed uniformly on a spherical surface (radius = 2.0 cm), and a second charge of −2.0 pC is distributed uniformly on a concentric spherical surface (radius = 4.0 cm). Determine the magnitude of the electric field 3.0 cm from the center of the two surfaces.
A) 30 N/C
B) 50 N/C
C) 40 N/C
D) 20 N/C
E) 70 N/C

Question

Quizplus · Accepted Answer

To find the electric field at a point 3.0 cm from the center, we use Gauss's law. The point is outside the inner sphere (radius = 2.0 cm) but inside the outer sphere (radius = 4.0 cm). Therefore, only the charge on the inner sphere contributes to the electric field at this point. The electric field E due to a point charge (or spherically symmetric charge distribution) at a distance r from the charge is given by Coulomb's law in the form derived from Gauss's law: $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$, where $Q$ is the charge, $r$ is the distance from the charge, and $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} C^2/N\cdot m^2$).Given:- $Q = 5.0 \, pC = 5.0 \times 10^{-12} \, C$- $r = 3.0 \, cm = 0.03 \, m$Plugging these values into the formula:$$E = \frac{1}{4\pi(8.85 \times 10^{-12})} \frac{5.0 \times 10^{-12}}{(0.03)^2}$$$$E = \frac{9 \times 10^9 \times 5.0 \times 10^{-12}}{0.03^2}$$$$E = \frac{45 \times 10^{-3}}{9 \times 10^{-4}}$$$$E = 50 \, N/C$$Thus, the magnitude of the electric field 3.0 cm from the center is 50 N/C.

A Charge of 5