At time t=0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet with speed 960 m/s . The sheet has uniform surface charge density 2.34.×10−9C/m2 . What is the speed of the proton at 5.40

Respuesta :

Explanation:

Formula for the electric field due to the infinite sheet of charge is as follows.

               E = [tex]\frac{\sigma}{2 \epsilon_{o}}[/tex]

where,   [tex]\sigma[/tex] = surface charge density

Now, formula for electric force acting on the proton is as follows.

             F = eE

where,    e = charge of the proton

According to the Newton's second law of motion, the net force acting on the proton is as follows.

                       F = ma

                 a = [tex]\frac{eE}{m}[/tex]

                    = [tex]\frac{e(\frac{\sigma}{2 \epsilon_{o}})}{m}[/tex]

                    = [tex]\frac{e \sigma}{2m \epsilon_{o}}[/tex]

According to the kinematic equation, speed of the proton in perpendicular direction is as follows.

              [tex]v_{f} = v_{i} + at[/tex]

                     = [tex](0 m/s) + \frac{e \sigma}{2 m \epsilon_{o}}t[/tex]

                     = [tex]\frac{1.6 \times 10^{-19}C \times 2.34 \times 10^{-9} C/m^{2} \times 5.40 \times 10^{-8}s}{2 \times (1.67 \times 10^{-27} kg)(8.85 \times 10^{-12} C^{2}/Nm^{2}}[/tex]

                     = 683.974 m/s

Hence, total speed of the proton is as follows.

                v' = [tex]\sqrt{(960 m/s)^{2} + (683.974 m/s)^{2}}[/tex]

                    = [tex]\sqrt{921600 + 467820.43}[/tex]

                    = [tex]\sqrt{1389420.43}[/tex]

                    = 1178.73 m/s

Therefore, we can conclude that speed of the proton is 1178.73 m/s.

The speed of the proton after the given time is 1178.75 m/s.

Speed of the proton:

The electric field due to a large sheet with surface charge density (σ) is given by:

E = σ/2ε₀

Therefore the force exerted on the proton will be:

F = qE

where q is the charge of the proton

ma = qE

so acceleration is:

a = qE/m

m is the mass of the proton

Let the sheet is in the horizontal direction, so the electric field and force will be in a vertical direction.

The initial speed of proton in vertical direction is zero.

from the third equation of motion, the vertical speed of proton:

v' = 0 + at

given that time t =  5.4 × 10⁻⁸s

v' = qEt/m

v' = [tex]\frac{1.6\times10^{-19}\times(2.34\times10^{-9}/2\times8.85\times10^{-12})\times5.4\times10^{-8}}{1.67\times10^{-27}}[/tex]

v' = 684 m/s

Given that the horizontal speed is 960 m/s.

The resultant speed of the proton is :

v = [tex]\sqrt{(960)^2+(684)^2}\;m/s[/tex]

v = 1178.75 m/s

Learn more about equations of motion:

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