A proton is placed in a uniform electric field of 2.75\times10^{3}\frac{N}{C}

Maaghu 2021-11-23 Answered
A proton is placed in a uniform electric field of \(\displaystyle{2.75}\times{10}^{{{3}}}{\frac{{{N}}}{{{C}}}}\) Calculate:
a) the magnitude of the electric force felt by the proton
b) the proton’s acceleration;
c) the proton’s speed after \(\displaystyle{1.00}\mu{s}\) in the field, assuming it starts from rest.

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Expert Answer

Cherry McCormick
Answered 2021-11-24 Author has 8705 answers

Step 1
We are given a uniform electric field \(\displaystyle{E}=\frac{{2}}{{75}}\times{10}^{{{3}}}{\frac{{{N}}}{{{C}}}}\) applied to a proton where its charge is \(\displaystyle{q}=+{1.6}\times{10}^{{-{19}}}\) C and mass \(\displaystyle{m}={1.67}\times{10}^{{-{27}}}{k}{g}\) (See Appendix F)
a) We are asked to calculate the magnitude of the electric force \(\displaystyle{\left|{F}\right|}\) exerted on the proton. As we are given E, we can use equation 21.4 to find F
1) \(\displaystyle{F}={\left|{q}\right|}{E}\)
Where q is the charge of the proton. Now we can plug our values for q and E into equation (1) to get F
\(\displaystyle{F}={\left|{q}\right|}{E}={\left({\left|+{16}\times{10}^{{-{19}}}{C}\right|}\right)}{\left({2.75}\times{10}^{{{3}}}{\frac{{{N}}}{{{C}}}}\right)}={4.4}\times{10}^{{-{16}}}{N}\)
As the charge of the particle is positive, the force has the same direction of the electric field.
Step 2
b) We want to calculate the proton's acceleration a. As the electric field exerted a force on the proton acquire kinetic energy due to this force, we can use Newton's second law to find the acceleration by knowing the mass of the proton from Appendix F, and use the law in the form.
2) \(\displaystyle{a}={\frac{{{F}}}{{{m}}}}\)
Now let use plug the values for F and m into equation (2) to get a of the proton
\(\displaystyle{a}={\frac{{{F}}}{{{m}}}}={\frac{{{4.4}\times{10}^{{-{16}}}{N}}}{{{1.67}\times^{{-{27}}}{k}{g}}}}={2.63}\times{10}^{{{11}}}{\frac{{{m}}}{{{s}^{{{2}}}}}}\)
Step 3
c) In this part we want to find the speed v of the proton after time \(\displaystyle{t}={1.00}\mu{s}\) Let us assume that the proton starts from the rest, and as the motion is from the rest and the electric field is one direction, the motion will be in one direction.
For a constant acceleration, the velocity in one direction could be found from Newton's laws of motion in the next relation
3) \(\displaystyle{v}={v}_{{{0}}}+{a}{t}\)
Where \(\displaystyle{v}_{{{0}}}\) is the speed from the rest and equals zero. Now we can plug our values for \(\displaystyle{v}_{{{0}}},\ {a}\) and t into equation (3) to get the speed v of the proton
\(\displaystyle{v}={v}_{{{0}}}+{a}{t}={0}+{\left({2.63}\times{10}^{{{11}}}{\frac{{{m}}}{{{s}^{{{2}}}}}}\right)}{\left({1.00}\times{10}^{{-{6}}}{s}\right)}={2.63}\times{10}^{{{5}}}{\frac{{{m}}}{{{s}}}}\)
As shown, the proton has a large speed due to the electric field
Answer:
a) \(\displaystyle{\left|{F}\right|}={4.4}\times{10}^{{-{16}}}{N}\)
b) \(\displaystyle{a}={2.63}\times{10}^{{{11}}}{\frac{{{m}}}{{{s}^{{{2}}}}}}\)
c) \(\displaystyle{v}={2.63}\times{10}^{{{5}}}{\frac{{{m}}}{{{s}}}}\)

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James Etheridge
Answered 2021-11-25 Author has 7972 answers
Step 1
\(\displaystyle{F}={\left|{q}\right|}{E}\)
Since the field is uniform, the force is constant and accelerates the charge.
The acceleration is constant and we can use kinematic constant acceleration equations to find the final speed.
A proton has charge \(\displaystyle+{e}\) and mass \(\displaystyle\frac{{1}}{{67}}\times{10}^{{-{27}}}{k}{g}\) a) \(\displaystyle{F}={\left({1.60}\times{10}^{{-{19}}}{C}\right)}{\left({2.75}\times{10}^{{{3}}}{\frac{{{N}}}{{{C}}}}\right)}={4.40}\times{10}^{{-{16}}}{N}\)
c) \(\displaystyle{v}_{{{x}}}={v}_{{{0}{x}}}+{a}_{{{x}}}{t}\) gives
\(\displaystyle{v}={\left({2.63}\times{10}^{{{11}}}{\frac{{{m}}}{{{s}^{{{2}}}}}}\right)}{\left({1.00}\times{10}^{{-{6}}}{s}\right)}={2.63}\times{10}^{{{5}}}{\frac{{{m}}}{{{s}}}}\)
b) \(\displaystyle{a}={\frac{{{F}}}{{{m}}}}={\frac{{{4.40}\times{10}^{{-{16}}}{N}}}{{{1.67}\times{10}^{{-{27}}}{k}{g}}}}={2.63}\times{10}^{{{11}}}{\frac{{{m}}}{{{s}^{{{2}}}}}}\)
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