Question

At time t=0t=0 a proton is a distance of 0.360 mm from a very large insulating sheet of charge and is moving parallel to the sheet with speed 990 m/sm/s . The sheet has uniform surface charge density 2.34.×10−9C/m22.34.×10−9C/m2 .

Answer 1

Answer:

$1.34 * 10^{3}$m/s

Explanation:

Parameters given:

distance of the proton form the insulating sheet = 0.360mm

speed of the proton, $v_{x}$ = 990m/s

Surface charge density, σ = 2.34 x $10^{-9}$ C/$m^{2}$

We need to calculate the speed at time, t = 7.0 * $10^{-8}$s.

We know that the proton is moving parallel to the sheet, hence, we can say it is moving in the x direction, with a speed $v_{x}$ on the axis.

The electric force acting on the proton moves in the y direction, so this means it is moving with velocity $v_{y}$ in the y axis.

Hence, the resultant velocity of the proton is given by:

$v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v_{x}$ = 990m/s from the question. We need to find $v_{y}$ and then the resultant velocity v.

Electric field is given in terms of surface charge density, σ as:

E = σ/ε0

where ε0 = permittivity of free space

=> $E = \frac{2.34 * 10^{-9} } {2 * 8.85418782 * 10^{-12} }$

E =  132 N/C

Electric Force, F is given in terms of Electric field:

F = eE

where e = electronic charge

=> F = ma = eE

∴ a = eE/m

where

a = acceleration of the proton

m = mass of proton

$a = \frac{1.60 * 10^{-19} * 132}{1.672 * 10^{-27} }$

a = 1.3 * $10^{10}$ m/$s^{2}$

Therefore, at time, t = 7.0 * $10^{-8}$, we can use one of the equations of linear motion to find the velocity in the y axis:

$a = \frac{v_{y} - v_{0}}{t} \\\\=> v_{y} = v_{0} + at$

$v_{y}$ = 0 + (1.3 * $10^{10}$ * 7.0 * $10^{-8}$)

$v_{y}$ = 910 m/s

∴ $v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v = \sqrt{990^{2} + 910^{2} }$

$v = \sqrt{1808200}$

v = 1344.69 m/s = $1.34 * 10^{3}$m/s