Skip to main content
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
모든 교과서Young & Freedman Calc 15th EditionCh 21: Electric Charge and Electric Field문제 38
21장, 문제 38

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?
Square with charges: +q at top corners, -q at bottom corners, side length a, dot at center representing net electric field.

검증된 단계별 안내
1
First, identify the configuration of the charges. The square has four corners, and each corner has a point charge. Two charges are positive (+q) and two are negative (-q). Arrange them such that opposite corners have the same type of charge.
Next, consider the symmetry of the problem. The center of the square is equidistant from all four charges. Due to symmetry, the electric field contributions from opposite charges will have components that cancel each other out.
Calculate the electric field due to a single charge at the center of the square. The distance from the center to any corner is \( \frac{a}{\sqrt{2}} \). Use the formula for the electric field due to a point charge: \( E = \frac{k \cdot q}{r^2} \), where \( k \) is Coulomb's constant and \( r \) is the distance from the charge to the point of interest.
Determine the direction of the electric field vectors. The electric field due to a positive charge points away from the charge, while the field due to a negative charge points towards the charge. Analyze the vector components along the x and y axes for each charge.
Finally, sum the vector components of the electric fields from all four charges to find the net electric field at the center. Due to symmetry, the net electric field will have a specific direction along one of the diagonals of the square. Calculate the magnitude using vector addition of the individual electric fields.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
12m

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 defined as E = F/q, where F is the force experienced by a test charge q. The direction of the electric field is the direction of the force on a positive test charge.
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 resultant field at a point by adding the contributions from each charge, considering both magnitude and direction.
추천 영상:
가이드 코스
03:32
Superposition of Sinusoidal Wave Functions

Symmetry in Electric Fields

Symmetry plays a crucial role in simplifying electric field calculations. In this problem, the square configuration and equal magnitude of charges create symmetrical electric field contributions. This symmetry can be used to determine the direction and simplify the calculation of the net electric field at the center of the square.
추천 영상:
가이드 코스
08:20
Electric Field Lines
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.

6065
views
교과서 질문

Point charges q1=−4.5q_1=-4.5 nC and q2=+4.5q_2=+4.5 nC are separated by 3.13.1 mm, forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle of 36.936.9° with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.2×10−97.2\(\times\)10^{-9} Nm?

2827
views
2
rank
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.

3214
views
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?

3310
views
교과서 질문

A very long, straight wire has charge per unit length 3.20×10−103.20\(\times\)10^{-10} C/m. At what distance from the wire is the electric field magnitude equal to 2.502.50 N/C?

3899
views
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 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.

4118
views