The electric field strength E 0 E 0 is measured at a perpendicular distance R R from an infinitely large, thin sheet that contains a uniform positive charge density σ . σ. In terms of the stated properties and the permittivity of free space ε 0 , ε0, write an expression for the electric field strength E E at a perpendicular distance 2 R 2R from the sheet.

same value in R and 2R       E = E₀ = σ / 2ε₀

Explanation:

For this exercise we use Gauss's law

            Ф = E. dA = /ε₀

We define a Gaussian surface with a cylinder with the base being parallel to the load sheet, so the electic field line and the normal line to the base are parallel and the scalar product is reduced to the algebraic product, in the parts the angle is 90º and the dot product is zero

As the sheet has two faces

           2E A = q_{int} /ε₀

The charge inside the cylinder is

           σ = q_{int} / A

           q_{int} = σ A

We substitute

           E = σ / 2ε₀

We see that this expression is independent of the distance, so it has the same value in R and 2R

             E = E₀ = σ / 2ε₀

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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