Gauss's law you are in the presence of a uniform electric field of strength e that is pointed in the x direction. you have 2 gaussian surfaces; one is a square of side length l, and the other is a circle of diameter l. which has the larger electric flux value?

the flow is the same on both surfaces

Explanation:

Gauss's law says that the electric field flow through a Gaussian surface is equal to the charge inside the surface divided by the electric permittivity constant, the equation is

          Φ =∫ E . dA = q / ε₀

The point is the scalar product and the bold ones indicate vector,

Let us examine the situations that give us, the elective field is uniform so that the field lines that cross the surface of the square are constant and perpendicular to two sides, therefore in flow on one side it is equal to the flow on the other side, but of negative sign; so the net flow is zero, in the two sides there are 90º between the field lines and the normal to the surface (dA) so the flow is zero

Now let's analyze the circular surface in this case the angles between the field that is uniform, lines are parallel, and the normal circle changes at each point, but by the symmetry of the circle the angle that there is an area at the top is the same of the surface at the bottom connected by a diameter, the net flow in each pair of points is canceled and consequently in flow throughout the circle is zero.

  We can also see the same result if we see that within each surface there are no charges, so the right of the Zero Gas Law

From the above described the two surfaces have zero flow

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s