A point charge q_1=+2.40 μC is held stationary at the origin. A s...

Question

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

Accepted Answer

Hey everyone. So this problem is about work and potential energy of particles, let's see what it is giving us and what it's asking of us. So we know that there is a stationary particle of a given charge. We'll call that Q one and that is 3. micro columns which we can rewrite as 10 to the minus six columns. And it is located at the origin of an Xy plane. So let's remember that an origin is just the zero on a plane. The position of a second particle were given that charge. So Q two is negative 4.40. Again, times 10 to the minus six columns and that second particle moves from one point to another point. It changes um in both the X and Y positions. And it's asking us to calculate the work done by the electric force on the second particle. So the first thing that we need to do is recall that the work done um by the electric force is that equation is given as work equals minus delta U. Where U. Is the electric potential energy. And so we can rewrite that as the opposite, so minus. It's a delta U. Is U f minus you not. Are you original? The beginning of the equation? And let's just clean up the negative signs here and that becomes you not plus you're not minus you. So we know that. We can also recall that the potential energy is given as K times Q one Q two over R Where K is our columns constant. So let's recall that that equals 8.99 times 10 to the ninth. Newton meter squared per Coolum squared. Okay, so the charges are constant. Colmes constant is constant. So the only thing that's changing between UF and you not is the radius and the radio this r. Is the distance between the charged particles. So we know at the beginning the distance is 8.5 cm. So we can rewrite that as 0.085 m. And so now the the next thing we need to figure out is what our F. Is, what our final position is. Think back a little bit. Remember pythagorean theorem. And recall that the hypotenuse of this triangle of this triangle created by um moving from minus 8.5 to 10 in the X on the explain on the X axis and 0 to 12.5 centimeters on the Y axis is going to give us that actual distance between charged particles. Okay, so quickly we could do that. It's going to be um X squared plus y squared. We're gonna rewrite those um these two centimeters in terms of meters. So it's just gonna be 20.1 m squared plus 0.125 m squared, plug that in. And we get Our final distance is .16 m. We write that over here. Okay, so let's go back up to our work equation, plug in what we know from our potential energy equation. And we get that work equals K Q one Q two minus R. Not or over. Are not minus K Q one Q two over our. We can make this simplify this just a little bit by pulling out Those KQ one Q 2 terms one over R not minus one over R. F. Is going to be our work. We are going to plug all of this into our calculator. We already have all of these values written in our known list above. And so we can just kind of plug and chug from here. Alright, the two charges were 3. times 10 to the minus six. Cool. Um and negative 4.40 times 10 to the negative six. Cool. Um It's really important to remember your signs when we're working with charges and we are going to multiply all of that. Bye one over. Are not zero m minus one over R. F. Which we calculated as 10.16 m Plug all of that in. And we get our work is negative .710 jewels. Let's look at our answer choices. And that aligns with B -7. Jules. All right. That's all we have for this problem. We'll see you in the next video. Mm

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

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