Question

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 φ1 through the surface of radius R, the electric flux φ2 through the surface of radius 2R is____________.a. φ2 = 1/4φ1b. φ2 = 1/2φ1c. φ2 = φ1d. φ2 = 2φ1e. φ2 = 4φ1

Answer 1

The electric flux through concentric spherical surfaces of different radii surrounding the same central charge is equal due to Gauss's Law, thus Φ2 = Φ1.

Step-by-step explanation:

When comparing the electric flux Φ1 through the surface of radius R to the electric flux Φ2 through the surface of radius 2R that surround a positive point charge Q located at the center, Gauss's Law is pertinent. Gauss's Law states that the electric flux through a closed surface depends only on the charge enclosed by the surface, not on the size or shape of the surface. Thus, since both surfaces enclose the same charge Q, the electric flux through each surface must be equal, regardless of their radii. Therefore, we can conclude that Φ2 is equal to Φ1, making the correct answer (c) Φ2 = Φ1.

Answer 2

c.
`\phi_1=\phi2`

Step-by-step explanation:

the electric flux is given by the Gaussian's theorem:

`\int \vec{E}\cdot d\vec{S}=(Q)/(\epsilon_0)`

where Q is the net charge inside the surface and e0 is the dielectric permittivity of vacuum. Due to in both cases the electric flux is independt of the radius of the surface and only depends of the charge Q. We conclude that both fluxes are equal.

`\phi_1=\phi2`

answer: c

hope this helps!!