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_1. When the two spheres are 0.800 m apart, q_2, is moving toward q_1 with speed 22.0 m/s (Fig. E23.5). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of q_2 when the spheres are 0.400 m apart?

Video transcript

Hey, everyone today, we're dealing with the problem regarding kinetic energy and electric potential energy. So we're being told that a point charge with a charge of negative 2.8 micro columns has a fixed position, a tiny ball with a massive eight g is charged and has a charge of negative 6.2 micro columns and is being shot towards the point charge. When the ball is 75 cm away from the point charge, its speed is 17 m/s. With this information, we're being asked to find what the speed of the ball is when it is 35 cm from the point charge. For the sake of this problem, we're also being asked to treat the ball as a point charge as well ending Nora's wait. So we're just going to sort of Ignore this 8g. It's a red herring, But let's go ahead and look through this. So um we know that separated charges because we have two charges right now. We have uh Q1, right? We have, I'll just label it Q Which is negative 2.8 micro columns. And we also have capital Q which is our ball And it has a charge of negative 6. micro columns. We're going to convert this two columns because we want to stay with R S I units. So recall that one micro Cullum is equal to uh 10 to the negative sixth columns. So this will be negative negative 2.80 times 10 to the negative sixth columns. And similarly, this will be negative 6.20 times 10 to the negative six columns. But with these two charges, we know that two separated charges possess electric potential energy. And when they're in motion, because the ball is being shot towards the point charge, there is also kinetic energy because the ball is in motion, we have to deal with kinetic energy here. So we're dealing with the inter conversions of energies, right. So we have kinetic energy, inter conversion of energy or is just conservation of energy in this case. So our initial kinetic energy plus the electric potential energy must be equal to the kinetic energy or the sum of the kinetic energy and potential energy at the end of our system because um energy can neither be created nor destroyed, just converted. Which means that if we have a set amount of energy at the start of our system, then once we have reached completion or the specific point in time, we should also theoretically have the same amount of energy. But with this information, let's also recall a few formulas, right, recall that uh kinetic energy can be calculated as one half M V squared and potential energy or electric potential energy. In this case can be described as K which is the electric constant, cool um constant 99 times 10 to the ninth Newton meters squared over um even even meter squared over cool um multiplied by the two charges divided by R the distance between set charges. So if we set this into our um formula here we get and that's right. This, this can therefore be translated to 1/2 M one right or sorry him, the initial squared plus que que que over our initial Because that will be 75 cm. Let's make this an arrow for less confusion. This would be one half M the final squared that's K Q Q over her final recall that Our initial is cm And our final is 35 cm because it's being shot from a distance towards the point charge all to recall that one centimeter is equal to tend to the native second meters. And since we want these to be in meters will simply um converted by multiplying into one over Or excusing, excuse me, multiplying by 10 to the negative second. And we get 0.75 m and 0.35 m respectively. So we're being asked to find what is the speed of the ball when it is cm from the point chart. So we're essentially, we're being asked to find what is the final speed in this case at of the ball. Let me write that and rent. So rather than rearranging everything and solving directly for um the V F squared to make it a little simpler for us, let's just isolate this entire uh final kinetic energy portion. Let me use blue. So we're going to essentially isolate this term right here. So rearranging this and let's keep that in blue. I suppose we can say that one half M the final squared is equal to 1/2 the initial squared plus que que que over our initial minus K Q Q over our final, which can be further simplified, excuse me, it can be further simplified two, 1/2 in the initial squared. And we can factor out the K Q Q terms leaving us with One over our initial -1 over our final. So that in mind if we go ahead and substitute in our values now into that equation, We get 1/2 in two. We're being told that the mass of the charge, excuse me, I made a mistake earlier, I said ignore the mass. The problem says ignore the weight, not the mass, which means we don't need to think about the effects of gravity here. But the ball still has charged my bad excuse my mistake. But we have the mass of 0. 08 kg. And we get this because we have eight g and we know that uh one kg is nothing but 10 to the negative three g, which gives us 0.8 kg. So that is our mass. The mass of the ball, the initial velocity was set to be 17 m per second meters per second plus Cool concept which is nine times 10 to the 9th Newton meters square over a cool um cool um squared multiplied by the two point charges negative 2.8 times 10 to the negative sixth columns. I'm going to continue it down here. Let me actually just move these over. In that case, we'll move these down here multiplied by Capital Q which is native 6.2 times 10 to the -6 columns multiplied by one over 0.75 m - over 0.35 m. So simplifying all of this, we get that the final kinetic energy, which is what this term is. Let me just write that this is kinetic energy. Final Will give us an answer of 0. jules. So if this is the final kinetic energy, right? And if this is equal to 1/2 m the final squared then and here I'll use red, then we can say that the final is equal to the square root of kinetic energy. Excuse me, two times kinetic energy, two times the kinetic energy divided by the mass. Which means if we equate this, we get two multiplied by 0.9179 jewels Divided by the mass, which is yet again, 0.008 kilograms, Which gives us an answer of 15 hoops, 15.1 m/s. Therefore, The speed of the ball when it is 35 cm from the point charge will be, excuse me, it's a choice. See, 15.1 m/s. I hope this helps and I look forward to seeing you all in the next one.