A point charge q1=+2.40 μC is held stationary at the origin. A second point charge q2=-4.30 μC m

step 1 In this question, we have two different charges. One charge Q1 is at the origin, and it carries

A point charge q1=+2.40 μC is held stationary at the origin....

A point charge $q_{1}=+2.40 \mu C$ is held stationary at the origin. A second point charge $q_{2}=-4.30 \mu C$ moves from the point $x=0.150 \mathrm{m}, y=0,$ to the point $x=0.250 \mathrm{m},$ $y=0.250 \mathrm{m} .$ How much work is done by the electric forceon $q_{2} ?$

Key Concepts

Electric Potential Energy

This concept refers to the energy stored in a system of charges due to their positions relative to one another. For point charges, it is typically calculated using Coulomb's law, where the potential energy depends on the product of the charges and the inverse of the distance between them. Understanding electric potential energy is essential in analyzing how energy changes when the configuration of charges is altered.

Work Done by Conservative Forces

In the context of electrostatics, the electric force is conservative, meaning that the work done by this force when moving a charge between two points depends only on the initial and final positions, not on the path taken. This property allows us to compute the work done by the electric force by evaluating the negative change in electric potential energy between the two positions.

Coulomb's Law

Coulomb's law describes the force between two point charges and is fundamental to calculating both the electric force and the potential energy in systems involving point charges. It establishes that the magnitude of the force is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them, providing the basis for further analysis of electric potential energy and work in an electrostatic field.

Transcript

00:01 In this question, we have two different charges.

00:02 One charge q1 is at the origin, and it carries a charge of q1 equals 2 .4.

00:12 And we have a second charge, q2, that is initially at the coordinate 0 .15 meter, x equals 0 .15, 0 equals 1 equals 0 and then it moves to a different distance it moves to x equals 0 .25 and 1 equals 0 .25 and we want to know how much work is done by the electric force on q2 this can be done by introducing the potential caused by one charge onto a separate charge so in chapter 18 section 1 we learn that the potential of a single charge is u equals k q1 q2 over r where r is a distance between q1 and q2 so what we can do is we can find if this is a position a and this is a position b we can find the potential when q2 is at position a and we can find the potential when q2 is at position b and we know that the work done by the electric field is just u a minus ub.

01:29 So when we look at ua and ub, the only quantities that's going to change is r.

01:33 It's a distance between q1 and q2.

01:37 So all we need to do is we plug in q1, q2, and k.

01:43 K is just a constant.

01:44 We know all these three numbers and we calculate the potential at point a by plugging r between q1 and q2, which is 0 .15 meter, and then we do the same thing for ub and we subtract them.

01:59 So let us try to figure out this.

02:04 U a minus u b equals k q1 q2 over 1 over r a minus 1 over r b so now we can plug in all the numbers k equals 9 .0 times 10 to the 9 okay q1 is 2 .40 times 2 .40 times 10 to negative 6.

02:37 Q2 is negative 4 .30 times 10 to the negative 6.

02:51 And we keep doing 1 over.

02:54 R .a is 0 .15 meter...

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