Just How Strong Is the Electric Force? Suppose you had two small boxes, each containing 1.0 g of protons. (a) If one were placed on the moon by an astronaut and the other were left on the earth, and if they were connected by a very light (and very long!) string, what would be the tension in the string? Express your answer in newtons and in pounds. Do you need to take into account the gravitational forces of the earth and moon on the protons? Why? (b) What gravitational force would each box of protons exert on the other box?

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Hey, everybody today, we're dealing with the problem about Newton's law of gravity as well as with coolers lawn utilizing the two together in conjunction. So we're being told that the sun emits ions and electrons that travel through space, consider that 0.1 kg of electrons have just escaped the sun, an equal amount of electrons, an equal amount of electrons have arrived at Mars surface. In practice, reaching a planet's surface may be thwarted by the atmosphere of the planet's magnetic field. But in this case, we are assuming that it hasn't interfered. In this case with this information, we're being asked to find a few things. One, what is the net electric force that the two groups of electrons exert on each other? Assuming that the two sets of electrons are stationary and behave like point charges were also being asked if the gravitational force from the sun and Mars to be included in computing the net force or to find the net force. And if the gravitational force from the sun and Mars should be included in that as well as to find to find the gravitational force, the two sets of electrons exert on each other. So were only given a very little bit of information here. So we also have some values that we need to know that we can recall or that we can look up. So we need to know cool um Law first off cool Law states that the force between two point charges will be equal to the equilibrium or non equilibrium, constant cooling constant multiplied by the magnitudes of the two separate charges in question, Divided by the distance between the charges raised to the power of two. We're also going to need to utilize Newton's law of gravity which is very similar, excuse me, two coolers constant where G is a gravitational constant M one and M two are masses of two planetary objects and are as again, the distance between set objects raise the second power. We also need to know the mass of an electron. The mass Of an electron is equal to 1.6 times 10 to the negative 19th Oops that's number 10 to NATO 19th columns. And that the charge or excuse me, That is the charge the mass. My pet is 9.1 times 10 to the negative 31st kg. And the charge of an electron charge of an electron is 1. times 10 to the negative 19th columns. There we go. After this, we also have a few other values that we need to know for our calculations coming up. We know that the mass of the sun or we can recall or look up mass of the sun is 1.99 times 10 to the 30th kg. Mass of Mom Mars is equal to 6.4, 2 times 10 to the 23rd kilograms. The um with of the sun is um excuse me, 6.9, 6 times 10 to the Um 8th, sorry, not the with the radius of the Sun. I should say That's 10 to the eight m. And similarly, for Mars, we can say that it is 3.3, 9 times 10 to the 6th m. Now finally, Mars's orbital radius, Mars's orbital radius. Mars oh Rad, I will say that is 3.39, 3.39 times. Excuse me, wrong value 2.2, 8 times 10 to the 11th m. So with all this in mind, let's go ahead and look at the first part. What is the net electric force that the two groups of electrons exert on each other? Oops, I'm gonna scroll down a bit And we'll start with question one. So to do so we know that the orbital radius is measured from the center of the sun in Mars, right? So there's separation, the separation will be equal to Mars is or blue, it ran minus the some of the radius is of the sun and Mars. So expanding that and substituting in our values, we get 2.28 times 10 to the 11th meters minus, we have 6.9, 6 times 10 to the 8th m plus Uh 3.3, 9 Times 10 to the six m giving us an answer of 2.27, 3 times 10 to the 11th m. Now, with this answer, this gives us our our value for the excuse me for the, again, the two groups between the two groups from the center of the sun all the way to the center of Mars. This gives us the separation between them. So from here, we can utilize cool um Law School um Law to go ahead and find the net force between the two electrons or the two point charges in question. However, we don't know what the charge of our electrons is, right? We don't know what Q one and Q two, R Q one and Q two are equal to each other because they have the same mass. But We don't know what Q one or Q two is. And this can be amended really easily. See, we know that we have 0.001 kg of electrons and we know that one electron electron Is 9.1 times 10 to the Native kg. So if we calculate this, Calculating, this gives us excuse me, our kg will cancel out and we will get 1.1 Times Times 10 to the 27 electrons. However, we know that we need to find the charge of this many electrons, which can also be done fairly simply because your coal, The charge of one electron is 1.6 times 10 to the negative 19th Coolum for every one electron. So solving for that, we get 1.758 times 10 to the eighth cool. Um so that will be equal to Main set of Q one Which will be equal to the magnitude of Q two. So substituting that back into our equation, We then get nine times 10 to the 9th multiplied by groups will be Newton meters squared per Coolum been cool M squared multiplied by Q one times Q two, which is simply 1.758 times 10 to the yes To the 8th Coolum Square over. Let me just this over a bit. So you don't get confused, but then it'll be over the separation that we calculated earlier 2.273 times 10 to the 11th meters squared, Giving us an answer of 5.3, 8 Times to the 3rd Newtons. So that'll be our first answer. The net force between the two points of where the electrons are hitting. Our next question is asking us to find whether or not the forces due to gravity um are by the sun. And mars should be included in the calculations to find the total force on each set of electrons. So the way we can do that, let me scroll down a little bit, scroll down a bit the way we can do this simply by utilizing Newton's second law or not Newton's second law, but Newton's law of gravitation And let's do it for the sun 1st. So remember is G M one M 2 over R squared, excuse me. So G will be the gravitational constant which is 6.673 times 10 to the negative 11th, Newton times um meters squared over kilogram squared multiplied by The mass of the sun which is 1.99 times 10 to the 30th kg Multiplied by the mass of the electrons which is zero points or kg divided by 6.9, 6 times 10 to the 8th meters square, which is the radius of the sun, which will give us an answer of 0.274 Newtons. Similarly, we can do the same thing for Mars gravitation of Mars will follow the same suit The gravitational constant being 6.67, 3 times 10 to the night of 11th Newton meters squared per kilogram squared multiplied by 6.4, 2 times 10 to the 23rd kg. As you mentioned earlier is the mass of Mars, the mass of the electrons Mars is putting out Or Mars are receiving in this case is also 0.001 kg And its radius is 3.3, 9 times 10 to the 6th meters Hold squared which gives us an answer of 3.3 or sorry, 3.7, 3 times 10 To the negative 3rd newtons. So let's highlight those. Now we have the forces exerted by the sun and by Mars, right? But let's compare this to the forces experienced or the net force experienced. Excuse me, the net force that was experienced or net electric force that the two electrons are exerting upon each other that we calculated in step one, right. So that is 5.3, 8 times 10 to the third. So we can approximate that to about let's say 5480 Newtons. So compared to the force to, to the sun, it's far greater than 0.274 newtons. Similarly, it is far greater is also far, far greater than 3.73 times 10 to the negative third newtons, which is also 0.373 newtons. Therefore, therefore, the force due to gravity of the planets is negligible and shouldn't be used is negligible to the force on each set of electrons. So we don't need to include it in our calculations or we don't need to use it to contribute to our calculations. Rather. Finally, our last step, the last thing that we're being asked to do is to calculate the gravitational force the two sets of electrons actually exert on each other, not just the electric force like we calculated earlier. So again, and let me scroll down just a wee bit more we're going to need or Newton's gravitational law. So we know we have key the force the gravitational force exerted by each set of electrons on each other will be excuse me, 6.6, 7, 3 times 10 To the native 11 Newton meters squared over a kilogram squared. That is the gravitational constant. The mass of both electrons or electron groups is 0. kg. So 0.001 kg and the separation between them as we measured earlier is two point 2, 7, 3 times 10 to the 11th m. And that'll be squared which will give us a final answer of 1.29 times 10 To the -39th Newtons. I like that. So with all this in mind, if we look back at our question, our questions stem the net force. The net electric force that the two electron groups exerted on each other was 5.38 times 10 to the third newtons which eliminates answer choices A and C the excuse me, the gravitation force from the sun and Mars should not have been included in the compute computation of the net force on the electron groups and the gravitational force of the two sets of electrons. That is that the electrons exert on each other. My bad is 1.29 times 10 to the negative 39th newtons or answer choice B I hope this helps. And I look forward to seeing you all in the next one.

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Key Concepts

Electric Force

Gravitational Force

Tension in a String