19. two concentric, imaginary spherical surfaces of radius (r) and (2r) respectively, surround a positive…

19. two concentric, imaginary spherical surfaces of radius (r) and (2r) respectively, surround a positive point - charge (+q) located at the center of the surfaces. when compared to the electric flux (varphi_1) through the surface of radius (r), the electric flux (varphi_2) through the surface of radius (2r) is:\na) (varphi_2=\frac{1}{4}varphi_1)\nb) (varphi_2=\frac{1}{2}varphi_1)\nc) (varphi_2 = varphi_1)\nd) (varphi_2 = 2varphi_1)\ne) (varphi_2 = 4varphi_1)

19. two concentric, imaginary spherical surfaces of radius (r) and (2r) respectively, surround a positive point - charge (+q) located at the center of the surfaces. when compared to the electric flux (varphi_1) through the surface of radius (r), the electric flux (varphi_2) through the surface of radius (2r) is:\na) (varphi_2=\frac{1}{4}varphi_1)\nb) (varphi_2=\frac{1}{2}varphi_1)\nc) (varphi_2 = varphi_1)\nd) (varphi_2 = 2varphi_1)\ne) (varphi_2 = 4varphi_1)

Answer

Explanation:

Step1: Apply Gauss's law

Gauss's law states that $\varPhi_E=\frac{Q_{enclosed}}{\epsilon_0}$, where $\varPhi_E$ is the electric - flux through a closed surface and $Q_{enclosed}$ is the charge enclosed by the surface.

Step2: Determine the enclosed charge for each surface

Let the charge at the center be $-Q$. For the surface of radius $R$, the charge enclosed $Q_{1}=-Q$. So, $\varPhi_1 = \frac{-Q}{\epsilon_0}$. For the surface of radius $2R$, the charge enclosed $Q_{2}=-Q$ (since the charge is located at the center and the Gaussian surfaces are concentric). So, $\varPhi_2=\frac{-Q}{\epsilon_0}$.

Answer:

C. $\varPhi_2=\varPhi_1$