A proton is located in a uniform electric field of&nbsp;2.75×103NC&nbsp;Find:&nbsp;a) the magnitude of the electric force felt by the proton&nbsp;b) the proton’s acceleration;&nbsp;c) the proton’s speed after&nbsp;1.00μs&nbsp;in the field, assuming it starts from rest.

Question

A proton is located in a uniform electric field of&amp;nbsp;2.75×103NC&amp;nbsp;Find:&amp;nbsp;a) the magnitude of the electric force felt by the proton&amp;nbsp;b) the proton’s acceleration;&amp;nbsp;c) the proton’s speed after&amp;nbsp;1.00μs&amp;nbsp;in the field, assuming it starts from rest.

Cherry McCormick · Accepted Answer

Step 1 We are given a uniform electric field E=275×103NC applied to a proton where its charge is q=+1.6×10−19 C and mass m=1.67×10−27kg (See Appendix F) a) We are asked to calculate the magnitude of the electric force |F| exerted on the proton. As we are given E, we can use equation 21.4 to find F 1) F=|q|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 F=|q|E=(|+16×10−19C|)(2.75×103NC)=4.4×10−16N 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&#039;s acceleration a. As the electric field exerted a force on the proton acquire kinetic energy due to this force, we can use Newton&#039;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) a=Fm Now let use plug the values for F and m into equation (2) to get a of the proton a=Fm=4.4×10−16N1.67×−27kg=2.63×1011ms2 Step 3 c) In this part we want to find the speed v of the proton after time t=1.00μs 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&#039;s laws of motion in the next relation 3) v=v0+at Where v0 is the speed from the rest and equals zero. Now we can plug our values for v0, a and t into equation (3) to get the speed v of the proton v=v0+at=0+(2.63×1011ms2)(1.00×10−6s)=2.63×105ms As shown, the proton has a large speed due to the electric field Answer: a) |F|=4.4×10−16N b) a=2.63×1011ms2 c) v=2.63×105ms

James Etheridge · Accepted Answer

Step 1  F=|q|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 +e and mass 167×10−27kg a) F=(1.60×10−19C)(2.75×103NC)=4.40×10−16N  c) vx=v0x+axt gives  v=(2.63×1011ms2)(1.00×10−6s)=2.63×105ms  b) a=Fm=4.40×10−16N1.67×10−27kg=2.63×1011ms2

Don Sumner · Accepted Answer

Answer:a) 4.4×10−16 Nb) 2.63×1011 m/s²c) [v≈725m/s]Explanation:a) To find the magnitude of the electric force felt by the proton, we can use the equation F=qE, where F is the force, q is the charge of the proton, and E is the electric field.The charge of a proton is 1.6×10−19 C, and the electric field is given as 2.75×103 N/C. Substituting these values into the equation, we have:[F=(1.6×10−19C)(2.75×103N/C)]Calculating this product, we find:[F=4.4×10−16N]Therefore, the magnitude of the electric force felt by the proton is 4.4×10−16 N.b) The acceleration of the proton can be determined using Newton&#039;s second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration (F=ma). Since the mass of a proton is approximately 1.67×10−27 kg, we can rearrange the equation to solve for acceleration:[a=Fm]Substituting the known values, we have:[a=4.4×10−16N1.67×10−27kg]Evaluating this expression, we find:[a≈2.63×1011m/s2]Therefore, the proton&#039;s acceleration is approximately 2.63×1011 m/s².c) To determine the proton&#039;s speed after it has traveled a distance of 1.00 µm in the electric field, assuming it starts from rest, we can use the equation of motion:[v2=u2+2as]where v is the final velocity, u is the initial velocity (which is 0 m/s since it starts from rest), a is the acceleration, and s is the distance traveled.Plugging in the values, we have:[v2=(0m/s)2+2×(2.63×1011m/s2)×(1.00×10−6m)]Simplifying this expression, we get:[v2=5.26×105m2/s2]Taking the square root of both sides, we find:[v≈725m/s]Therefore, the proton&#039;s speed after traveling 1.00 µm in the electric field is approximately 725 m/s.

RizerMix · Accepted Answer

a) To find the magnitude of the electric force felt by the proton, we can use the equation F=qE, where F is the force, q is the charge, and E is the electric field strength. For a proton, the charge is 1.6×10−19 C. Plugging in the values, we have:[F=(1.6×10−19C)×(2.75×103N/C)]Calculating this expression gives us the magnitude of the electric force felt by the proton.b) The acceleration of the proton can be calculated using Newton&#039;s second law, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. Since the proton has a very small mass of 1.67×10−27 kg, we can consider it negligible compared to the force experienced. Therefore, the acceleration of the proton is given by:[a=Fm=(1.6×10−19C)×(2.75×103N/C)1.67×10−27kg]Evaluating this expression gives us the proton&#039;s acceleration.c) To find the proton&#039;s speed after 1.00 μs in the electric field, assuming it starts from rest, we can use the equation v=u+at, where v is the final velocity, u is the initial velocity (which is 0 m/s in this case), a is the acceleration, and t is the time. Plugging in the values, we have:[v=0+(a×1.00×10−6s)]Evaluating this expression gives us the proton&#039;s speed after 1.00 μs in the electric field.

karton · Accepted Answer

a) We can use the formula to determine the size of the electric force that the proton experiences F=qE, where F is the force, q is the charge of the proton, and E is the electric field strength.The charge of a proton, q, is equal to the elementary charge, e, which is approximately 1.602×10−19 C. The electric field strength, E, is given as 2.75×103 N/C.Substituting these values into the formula, we get:F=(1.602×10−19C)(2.75×103N/C)Therefore, the magnitude of the electric force felt by the proton is 4.405×10−16 N.b) The acceleration of the proton can be determined using Newton&#039;s second law of motion, which states that the force experienced by an object is equal to its mass multiplied by its acceleration, F=ma.Since the mass of a proton is approximately 1.673×10−27 kg and we have already found the force to be 4.405×10−16 N, we can rearrange the formula to solve for the acceleration:a=Fm=4.405×10−16N1.673×10−27kgAfter performing the calculation, we find that the proton&#039;s acceleration is 2.631×1011 m/s2.c) To determine the proton&#039;s speed after traveling for 1.00 μs (microseconds) in the electric field, assuming it starts from rest, we can use the equation of motion:v=u+atwhere v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration, and t is the time.Plugging in the values, we get:v=0+(2.631×1011m/s2)×(1.00×10−6s)Simplifying the equation, we find that the proton&#039;s speed after 1.00 μs in the electric field is 2.631×105 m/s.

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

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