A proton is released in a uniform electric field, and it experiences an electric force of&nbsp;3.75×10−14&nbsp;N&nbsp;toward the south. What are the magnitude and direction of the electricfield? The equation is E=f/q, but there is only one number and three variables. What numbers go in the other two?

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A proton is released in a uniform electric field, and it experiences an electric force of&amp;nbsp;3.75×10−14&amp;nbsp;N&amp;nbsp;toward the south. What are the magnitude and direction of the electricfield? The equation is E=f/q, but there is only one number and three variables. What numbers go in the other two?

Mayme · Accepted Answer

The q in the equation is the charge on a proton which is  1.6×10−19 C  and F (is given)  =3.75×10−14 N  So solve for E and you should get that the positive charge is moving south therefore the electric field is also pointing south

Eliza Beth13 · Accepted Answer

Given that the electric force experienced by the proton is 3.75×10−14N toward the south, we can substitute this value into the equation and solve for the electric field.Let&#039;s define the magnitude of the electric field as E and the direction as θ.𝐄=𝐅qE=Fq  [Since we&#039;re only looking for the magnitude, we can ignore the direction temporarily.]Now, substituting the given value of the electric force, we have:E=3.75×10−14NqHowever, we need to determine the value of the charge, q, in order to find the electric field magnitude. The charge of a proton is 1.6×10−19C. Substituting this value into the equation:E=3.75×10−14N1.6×10−19CTo simplify the expression, we can divide the numerator and denominator by 10−19:E=3.75×10−14N1.6×10−19C×110−19E=3.75×10−14N1.6E=2.34375×10−14N/CTherefore, the magnitude of the electric field is 2.34375×10−14N/C.Now let&#039;s determine the direction of the electric field. We were given that the electric force experienced by the proton is toward the south. In the context of electric fields, the direction of the electric field is the direction in which a positive charge would experience a force. Since protons are positively charged, the electric field must point in the opposite direction of the force experienced by the proton.Hence, the direction of the electric field is north.In summary, the magnitude of the electric field is 2.34375×10−14N/C, and the direction is north.

madeleinejames20 · Accepted Answer

Step 1. Given information:- Electric force (F) on the proton: F=3.75×10−14N (toward the south).Step 2. Formula:- Electric field (E) is given by the equation: E=Fq, where q represents the charge.Step 3. Solution:- We have the value for the force (F), but we need to find the magnitude and direction of the electric field (E).- Since the proton has a positive charge, we know that the direction of the electric field will be opposite to the direction of the force experienced by the proton.Step 4. Calculation:- Let&#039;s assume the charge of the proton (q) is 1.6×10−19C (the charge of a proton).- Substituting the values into the formula, we have:E=3.75×10−14N1.6×10−19CStep 5. Simplification:- We can simplify the expression by dividing the numerator and denominator by 10−19:E=3.751.6×10−1410−19V/mStep 6. Further simplification:- Performing the division, we get:E=2.34375×105V/mStep 7. Conclusion:- The magnitude of the electric field is 2.34375×105V/m.- The direction of the electric field is toward the north (opposite to the direction of the force experienced by the proton).

Nick Camelot · Accepted Answer

Answer:2.34375×105N/CExplanation:Given:Electric force, f=3.75×10−14N toward the south.Charge of a proton, q=1.6×10−19C (This value is a constant and represents the charge of a proton.)We are asked to find the magnitude and direction of the electric field.Using the equation E=fq, we can substitute the known values:E=3.75×10−14N1.6×10−19CSimplifying this expression gives:E=2.34375×105N/CSo the magnitude of the electric field is 2.34375×105N/C.To determine the direction, we refer to the information given in the problem: the electric force is toward the south. The direction of the electric field is defined as the direction a positive charge would experience a force. Since the proton is positively charged, it experiences a force toward the south. Therefore, the direction of the electric field is also toward the south.Thus, the magnitude of the electric field is 2.34375×105N/C and its direction is south.

A proton is released in a uniform electric field, and it experiences an electric force of 3.75\times10^{-14}\ N toward the south. What are the magnitude and direction of the electricfield? The equation is E=f/q, but there is only one number and three variables. What numbers go in the other two?

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