Gaussian surface 1 has twice the area of Gaussian surface 2. Both surfaces enclose the same charge Q. Is the electric flux through surface 1 greater than, less than, or the same as the electric flux through surface 2?

The electric flux through surface 1 is greater than the electric flux through surface 2.

Step-by-step explanation:

Electric Flux is given as:

Φ = E*A

where E = Electric field

A = surface area.

For surface 1,

Φ(1) = E(1)*A(1)

For surface 2,

Φ(2) = E(2)*A(2)

Since both surfaces have the same charge, the electric fields will be the same.

Hence, E(1) = E(2) = E

From the question we are given that:

A(1) = 0.5A(2)

Hence,

Φ(1) = E*A(1)

Φ(2) = E*A(2) = E*0.5A(1)

Φ(2) = 0.5E*A(1)

Comparing Φ(1) and Φ(2), we see that Φ(1) is 2 time greater than Φ(2), hence, the Flux through surface 1 is greater than the Flux through surface 2.

the answer would be d.

step-by-step explanation:

move all variables to the left side and all constants to the right side

(x−3)^2+(y−1)^2=16

this is the form of a circle. use this form to determine the center and radius of the circle.

( x − h ) 2 + ( y − k ) 2 = r 2

match the values in this circle to those of the standard form. the variable   r   represents the radius of the circle,   h   represents the x-offset from the origin, and   k   represents the y-offset from origin.

r = 4

h = 3

k = 1

the center of the circle is found at   ( h , k ) .

center:   ( 3 , 1 )

these values represent the important values for graphing and analyzing a circle.

center:   ( 3 , 1 ).

radius:   4