A small particle has charge -5.00 μC and mass 2.00x10^-4 kg. It m...

Question

A small particle has charge -5.00 μC and mass 2.00x10^-4 kg. It moves from point A, where the electric po-tential is V_A = +200 V, to point B, where the electric potential is V_B = +800 V. The electric force is the only force acting on the particle. The particle has speed 5.00 m/s at point A. What is its speed at point B? Is it moving faster or slower at B than at A? Explain.

Accepted Answer

Hey everyone in this problem. We have a test charge that carries a charge of negative 7. micro cool apps and has a mass of 0.24 g. Okay, It moves from point P to point Q. The potential at P and Q. Is 100 and 25 volts and 584 volts respectively. Okay. If the net force on the test charge is the electric force and the charge has a speed of 2.4 m per second at point P. Were asked to calculate its speed at point Q. Okay. Alright. So in this case were given one speed. Were asked for a speed at another point. Okay. And here we have mechanical energy conserved. Okay, so let's recall. Our conservation of mechanical energy gives us that the kinetic energy of point P plus the potential energy at point P is going to equal to the kinetic energy at point Q plus the potential energy at point Q. And we have conservative forces only. And so we get this conservation of mechanical energy. Now recall kinetic energy is gonna be one half M B squared. So our K. P. is gonna be 1/2 mm. Okay. And the mass of the test charges the same when it moves. Okay, so we have the same mass both at point P and point Q. Okay, so we have one half M. V. P squared in our potential energy here we're talking about potential energy of it charge. Okay. And so this is going to be equal to Q. The charge times V. Big V the potential. And let's not mistake the V. S. Here we have little V. P. More talking kinetic energy, the speed. And we have big VP. When we're talking about the potential energy, that's the potential. Okay. Alright. Same thing on the right hand side we get one half M. V. Q squared plus Q. Q. Big V. Cute. And what we're looking for is the speed up point Q. So we're looking for this little V. Q. Now we know the mass. We know the speed at point P. We know the charge and we know the potential. So we can just go ahead, plug in the information we know and find saul for me. Okay, so we get one half Times The Max which is 0.24. Okay, now we're given 0.24 g. We want to convert this into our standard unit. So we have 0.24 Times 10 to the negative three kg V. P. Is our speed up. Point P. Which we're told is 2.4 meters per second. All squared plus que the charge of negative 7.2. Okay, here we have micro columns. We want to put this into again, standard unit of Cool. Um So we have times 10 to the negative six columns Times the potential at Point P. Which is 125V. Alright, that's the left hand side. On the right hand side. It's going to get a little messy. So I'm gonna put the equal sign underneath just for this line. Um so we have room to write it all up. We have one half Times a mass which is again . Times 10 to the negative three kg times V Q squared Plus the charge, negative 7.2 times 10 to the -6 columns times the potential at Q. Which is 584 folds. Alright, great. Now we have all of our values in here. It's just a matter of solving. So on the left hand side we are going to get 0.000 12. Okay, in our unit here, kilogram meter squared per second squared. That's a jewel. That's the first term. And then the second term is going to be negative 0.0009 jules. And this is going to be equal to 0. 012 telegrams times V q squared zero 048 jules. All right, great. Now we can simplify, Okay, we can move this over to the left hand side. So we have 0.6912 minus 0.9. And then we're gonna add this term on the right hand side. We're going to get 0. 3996 jules is equal to 0.12 kg times v q squared. Okay, we divide and we get that V Q r speed at point q squared is 33.3 meter squared per second squared. Ok? We get drool divided by kilogram. So we're left with meter squared per second squared. We take the square root. We get VQ Is equal to positive or negative 5.77 meters per second. Okay? And we're just asked for the speed. And so we just want to know the magnitude of that value. So we're just gonna take the positive route Of 5. m/s. Okay? There we go. So the speed is 5.77 m per second at point Q. That's gonna be answer. C Thanks everyone for watching. I hope this video helped see you in the next one.

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