# Answer correct magnitude and direction of electric

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ANSWER: Correct Magnitude and Direction of Electric Fields A small object A, electrically charged, creates an electric field. At a point P located 0.250 directly north of A, the field has a value of 40.0 directed to the south. Part A What is the charge of object A? Hint 1. How to approach the problem Recall that the electric field at a point P due to a point charge is proportional to the magnitude of the charge and inversely proportional to the square of the distance of P from the charge. Furthermore, the direction of the field is determined by the sign of the charge. Hint 2. Find an expression for the charge Which of the following expressions gives the correct magnitude of charge that produces an electric field of magnitude at a distance from the charge? In the following expressions is a constant that has units of . It is strongly repelled. It is strongly attracted. It is weakly attracted. It is weakly repelled. It is neither attracted nor repelled. m N/C q E r k N ⋅ / m 2 C 2

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Hint 1. Magnitude of the electric field of a point charge Given a point charge , the magnitude of the electric field at a distance from the charge is given by , where the constant of proportionality is = 8.99×10 9 . ANSWER: Hint 3. Find the sign of the charge What is the sign of the charge that produces an electric field that points toward the charge? ANSWER: ANSWER: Correct Part B q E r E = k | q | r 2 k N ⋅ / m 2 C 2 q = k E d 2 q = kEd 2 q = k d 2 E q = Ed 2 k positive negative 1.11×10 −9 −1.11×10 −9 2.78×10 −10 −2.78×10 −10 5.75×10 12 −5.75×10 12 C C C C C C
If a second object B with the same charge as A is placed at 0.250 south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P ? Hint 1. How to approach the problem Since the electric field is a vector quantity, you need to apply the principle of superposition to find the total field at P. The principle of superposition in terms of electric fields says that the total electric field at any point due to two or more charges is the vector sum of the fields that would be produced at that point by the individual charges. Hint 2. Find the vector sum of the electric fields Which of the following diagrams, where and are the electric fields produced by A and B, respectively, correctly represents the situation described in this problem? ANSWER: Hint 3. Find the electric field produced by B at P What is the magnitude of the electric field produced by the second object B at point P? Express your answer in newtons per coulomb. Hint 1. Magnitude of the electric field of a point charge Given a point charge , the magnitude of the electric field at a distance from the charge is given by , m E PA E PB A B C D E 2 q E r E = k | q | r 2

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where the constant of proportionality is = 8.99×10 9 .
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The text passage is a homework help document that includes a physics problem about electric fields, with a series of hints and questions guiding the student to the solution.
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First, describe the content structure you found in 1-2 sentences.

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TYPE: <one of the following values>
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# Answer correct magnitude and direction of electric

• Homework Help
• cee878
• 27
• 100% (3) 3 out of 3 people found this document helpful

This preview shows page 13 - 17 out of 27 pages.

ANSWER: Correct Magnitude and Direction of Electric Fields A small object A, electrically charged, creates an electric field. At a point P located 0.250 directly north of A, the field has a value of 40.0 directed to the south. Part A What is the charge of object A? Hint 1. How to approach the problem Recall that the electric field at a point P due to a point charge is proportional to the magnitude of the charge and inversely proportional to the square of the distance of P from the charge. Furthermore, the direction of the field is determined by the sign of the charge. Hint 2. Find an expression for the charge Which of the following expressions gives the correct magnitude of charge that produces an electric field of magnitude at a distance from the charge? In the following expressions is a constant that has units of . It is strongly repelled. It is strongly attracted. It is weakly attracted. It is weakly repelled. It is neither attracted nor repelled. m N/C q E r k N ⋅ / m 2 C 2

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Hint 1. Magnitude of the electric field of a point charge Given a point charge , the magnitude of the electric field at a distance from the charge is given by , where the constant of proportionality is = 8.99×10 9 . ANSWER: Hint 3. Find the sign of the charge What is the sign of the charge that produces an electric field that points toward the charge? ANSWER: ANSWER: Correct Part B q E r E = k | q | r 2 k N ⋅ / m 2 C 2 q = k E d 2 q = kEd 2 q = k d 2 E q = Ed 2 k positive negative 1.11×10 −9 −1.11×10 −9 2.78×10 −10 −2.78×10 −10 5.75×10 12 −5.75×10 12 C C C C C C
If a second object B with the same charge as A is placed at 0.250 south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P ? Hint 1. How to approach the problem Since the electric field is a vector quantity, you need to apply the principle of superposition to find the total field at P. The principle of superposition in terms of electric fields says that the total electric field at any point due to two or more charges is the vector sum of the fields that would be produced at that point by the individual charges. Hint 2. Find the vector sum of the electric fields Which of the following diagrams, where and are the electric fields produced by A and B, respectively, correctly represents the situation described in this problem? ANSWER: Hint 3. Find the electric field produced by B at P What is the magnitude of the electric field produced by the second object B at point P? Express your answer in newtons per coulomb. Hint 1. Magnitude of the electric field of a point charge Given a point charge , the magnitude of the electric field at a distance from the charge is given by , m E PA E PB A B C D E 2 q E r E = k | q | r 2

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where the constant of proportionality is = 8.99×10 9 .
• Spring '14

{[ snackBarMessage ]}

###### "Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern
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The text passage is a homework help document that includes a physics problem about electric fields, with a series of hints and questions guiding the student to the solution.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS