A point charge q1=+2.40q_1=+2.40 μμC is held stationary at the origin. A second point charge q2=−4.30q_2=-4.30 μμC moves from the point x=0.150x=0.150 m, y=0y=0 to the point x=0.250x=0.250 m, y=0.250y=0.250 m. How much work is done by the electric force on q2q_2?

Ch 23: Electric Potential

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610당신이 사용하는 게 아니라요?교과서 변경

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Young & Freedman Calc 14th Edition

Ch 23: Electric Potential

문제 1

23장, 문제 1

A point charge $q sub 1 equals positive 2.40$ $μ$ C is held stationary at the origin. A second point charge $q_{2} = - 4.30$ $μ$ C moves from the point $x equals 0.150$ m, $y equals 0$ to the point $x equals 0.250$ m, $y equals 0.250$ m. How much work is done by the electric force on $q sub 2$ ?

검증된 단계별 안내

First, understand that the work done by the electric force is related to the change in electric potential energy as the charge q_2 moves from its initial position to its final position.

Calculate the initial electric potential energy (U_i) at the starting position (x=0.150 m, y=0) using the formula: U_i = (k * q_1 * q_2) / r_i, where k is Coulomb's constant (8.99 x 10^9 N m²/C²) and r_i is the initial distance between q_1 and q_2.

Determine the initial distance r_i between the charges using the distance formula: r_i = √((x_2 - x_1)^2 + (y_2 - y_1)^2), where (x_1, y_1) is the position of q_1 at the origin and (x_2, y_2) is the initial position of q_2.

Calculate the final electric potential energy (U_f) at the final position (x=0.250 m, y=0.250 m) using the formula: U_f = (k * q_1 * q_2) / r_f, where r_f is the final distance between q_1 and q_2.

Determine the final distance r_f using the distance formula: r_f = √((x_2 - x_1)^2 + (y_2 - y_1)^2), where (x_2, y_2) is the final position of q_2. The work done by the electric force is then given by the change in potential energy: Work = U_i - U_f.

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주요 개념

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

Electric Force

Electric force is the interaction between charged particles, described by Coulomb's law. It is proportional to the product of the charges and inversely proportional to the square of the distance between them. In this scenario, the force acts on q_2 due to q_1, influencing its movement and the work done.

Work Done by a Force

Work done by a force is the energy transferred when an object is moved by that force. It is calculated as the dot product of the force vector and the displacement vector. In this context, the work done by the electric force on q_2 is determined by the change in position and the force exerted by q_1.

Electric Potential Energy

Electric potential energy is the energy a charge possesses due to its position in an electric field. It depends on the charge and the electric potential at its location. The change in electric potential energy as q_2 moves from one point to another is directly related to the work done by the electric force.

관련 실천

교과서 질문

Two protons, starting several meters apart, are aimed directly at each other with speeds of $2.00 \times 10^{5}$ m/s, measured relative to the earth. Find the maximum electric force that these protons will exert on each other.

How much work would it take to push two protons very slowly from a separation of $2.00 \times 10^{- 10}$ m (a typical atomic distance) to $3.00 \times 10^{- 15}$ m (a typical nuclear distance)? If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation?

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교과서 질문

A small metal sphere, carrying a net charge of $q_{1} = - 2.80$ μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of $q_{2} = - 7.80$ μC and mass $1.50$ g, is projected toward $q sub 1$ . When the two spheres are $0.800$ m apart, $q sub 2$ , is moving toward $q sub 1$ with speed $22.0$ m/s (Fig. E $23.5$ ). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. What is the speed of $q sub 2$ when the spheres are $0.400$ m apart?

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