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Ch 21: Electric Charge and Electric Field
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552당신이 사용하는 게 아니라요?교과서 변경
21장, 문제 36b

A 4.00-4.00-nC point charge is at the origin, and a second 5.00-5.00-nC point charge is on the xx-axis at x=0.800x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at each point in part (a). Note: Part (a) asked to find the electric field (magnitude and direction) at each of the following points on the xx-axis: (i) x=0.200x = 0.200 m; (ii) x=1.20x = 1.20 m; (iii) x=0.200x = -0.200 m.

검증된 단계별 안내
1
To find the electric field at a point due to a point charge, use the formula: E = (k * |q|) / r^2, where E is the electric field, k is Coulomb's constant (8.99 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge to the point.
For point (i) x = 0.200 m, calculate the electric field due to each charge separately. For q1 at the origin, r = 0.200 m. For q2 at x = 0.800 m, r = 0.600 m. Determine the direction of each field: both charges are negative, so the electric field points towards the charges.
For point (ii) x = 1.20 m, calculate the electric field due to each charge. For q1, r = 1.20 m. For q2, r = 0.400 m. Again, determine the direction of each field: both fields point towards the charges.
For point (iii) x = -0.200 m, calculate the electric field due to each charge. For q1, r = 0.200 m. For q2, r = 1.000 m. Determine the direction of each field: both fields point towards the charges.
To find the net electric force on an electron at each point, use F = e * E, where e is the charge of an electron (1.60 x 10^-19 C). Calculate the force using the net electric field found at each point.

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이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
11m

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Electric Field

The electric field is a vector field around a charged particle that represents the force exerted per unit charge at any point in space. It is calculated using Coulomb's law, where the electric field E due to a point charge q at a distance r is given by E = k * |q| / r^2, with k being Coulomb's constant. The direction of the field is radially outward for positive charges and inward for negative charges.
추천 영상:
가이드 코스
03:16
Intro to Electric Fields

Superposition Principle

The superposition principle states that the net electric field due to multiple charges is the vector sum of the electric fields produced by each charge individually. This principle allows us to calculate the total electric field at a point by considering the contribution from each charge separately and then adding them together, taking into account both magnitude and direction.
추천 영상:
가이드 코스
03:32
Superposition of Sinusoidal Wave Functions

Electric Force on a Charge

The electric force on a charge in an electric field is given by F = qE, where F is the force, q is the charge, and E is the electric field at the location of the charge. For an electron, which has a negative charge, the direction of the force is opposite to the direction of the electric field. This concept is crucial for determining the net force on a charge placed in the vicinity of other charges.
추천 영상:
가이드 코스
05:37
Electric Charge
관련 실천
교과서 질문

A point charge q1=4.00q_1=-4.00 nC is at the point x=0.600x = 0.600 m, y=0.800y = 0.800 m, and a second point charge q2=+6.00q_2=+6.00 nC is at the point x=0.600x = 0.600 m, y=0y = 0. Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

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교과서 질문

A point charge is placed at each corner of a square with side length aa. All charges have magnitude qq. Two of the charges are positive and two are negative (Fig. E21.4221.42). What is the direction of the net electric field at the center of the square due to the four charges, and what is its magnitude in terms of qq and aa?

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교과서 질문

A 4.00-4.00-nC point charge is at the origin, and a second 5.00-5.00-nC point charge is on the xx-axis at x=0.800x = 0.800 m. Find the electric field (magnitude and direction) at each of the following points on the xx-axis: (i) x=0.200x = 0.200 m; (ii) x=1.20x = 1.20 m; (iii) x=0.200x = -0.200 m.

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교과서 질문

A +8.75+8.75-mC point charge is glued down on a horizontal frictionless table. It is tied to a 6.50-6.50-mC point charge by a light, nonconducting 2.502.50-cm wire. A uniform electric field of magnitude 1.85×1081.85\(\times\)10^8 N/CN/C is directed parallel to the wire, as shown in Fig. E21.3421.34. What would the tension be if both charges were negative?

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교과서 질문

A very long, straight wire has charge per unit length 3.20×10103.20\(\times\)10^{-10} C/m. At what distance from the wire is the electric field magnitude equal to 2.502.50 N/C?

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교과서 질문

A +8.75+8.75-mC point charge is glued down on a horizontal frictionless table. It is tied to a 6.50-6.50-mC point charge by a light, nonconducting 2.502.50-cm wire. A uniform electric field of magnitude 1.85×1081.85\(\times\)10^8 N/CN/C is directed parallel to the wire, as shown in Fig. E21.3421.34. Find the tension in the wire.

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