A metal sphere of radius R = 30 cm carries a total charge Q = 2.0 $\mu$C.What is the magnitude of the electric field at a point 5 m away? The point is located on the positive x-axis,where the center of the sphere is located at the origin.

Question

Quizplus · Accepted Answer

The electric field at a point outside a charged sphere can be calculated using the formula $E = \frac{kQ}{r^2}$, where $E$ is the electric field, $k$ is Coulomb's constant ($8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$), $Q$ is the total charge, and $r$ is the distance from the center of the sphere to the point. Given $Q = 2.0 \times 10^{-6} \, \text{C}$ and $r = 5 \, \text{m}$, substituting these values gives $E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-6})}{(5)^2} = 7.19 \times 10^{2} \, \text{N/C}$.

A Metal Sphere of Radius R = 30 Cm Carries μ\muμ

A Metal Sphere of Radius R = 30 Cm Carries $\mu$