Charge of uniform density (40 pC/m²) is distributed on a spherical surface (radius = 1.0 cm), and a second concentric spherical surface (radius = 3.0 cm) carries a uniform charge density of 60 pC/m². What is the magnitude of the electric field at a point 4.0 cm from the center of the two surfaces? A) 3.8 N/C B) 4.1 N/C C) 3.5 N/C D) 3.2 N/C E) 0.28 N/C

Question

Quizplus · Accepted Answer

To find the electric field at a point 4.0 cm from the center, we use Gauss's law. The total charge enclosed by a Gaussian surface at 4.0 cm includes both spherical surfaces. The charge on the first surface is $Q_1 = \sigma_1 \cdot 4\pi r_1^2 = 40 \times 10^{-12} \cdot 4\pi (0.01)^2$ C, and the charge on the second surface is $Q_2 = \sigma_2 \cdot 4\pi r_2^2 = 60 \times 10^{-12} \cdot 4\pi (0.03)^2$ C. The total charge $Q_{total} = Q_1 + Q_2$. The electric field at 4.0 cm, using $E = \frac{Q_{total}}{4\pi \epsilon_0 r^2}$, where $r = 0.04$ m and $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2$, gives a value close to 4.1 N/C, considering the total enclosed charge and the distance from the center.

Charge of Uniform Density (40 PC/m2) Is Distributed on a Spherical

Charge of Uniform Density (40 PC/m²) Is Distributed on a Spherical