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 .

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 [tex]v_{x}[/tex] on the axis.

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

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

[tex]v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}[/tex]

[tex]v_{x}[/tex] = 990m/s from the question. We need to find [tex]v_{y}[/tex] 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

=> [tex]E = \frac{2.34 * 10^{-9} } {2 * 8.85418782 * 10^{-12} }[/tex]

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

[tex]a = \frac{1.60 * 10^{-19} * 132}{1.672 * 10^{-27} }[/tex]

a = 1.3 * [tex]10^{10}[/tex] m/[tex]s^{2}[/tex]

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

[tex]a = \frac{v_{y} - v_{0}}{t} \\\\=> v_{y} = v_{0} + at[/tex]

[tex]v_{y}[/tex] = 0 + (1.3 * [tex]10^{10}[/tex] * 7.0 * [tex]10^{-8}[/tex])

[tex]v_{y}[/tex] = 910 m/s

∴ [tex]v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}[/tex]

[tex]v = \sqrt{990^{2} + 910^{2} }[/tex]

[tex]v = \sqrt{1808200}[/tex]

v = 1344.69 m/s = [tex]1.34 * 10^{3}[/tex]m/s