What is the magnitude of the net force on charge A in FIGURE EX22...

Question

What is the magnitude of the net force on charge A in FIGURE EX22.17?

Accepted Answer

Hi everyone in this practice problem, we're being asked to determine the net force acting on charge M as shown in the figure, we will have three different charges. Church M with 1.3 micro coom charge N with 2.1 Nano Colum and charge O with negative 1. micro Colum church N and O is located one centimeter apart and M and O is located three centimeter apart. Church M and N therefore, is going to be uh located two centimeters apart from one another. Since charge MN and O are going to be on the same axis. The options given for the net force acting on charge M is going to be a 18.2 Newton B 20.7 Newton C 24.3 Newton and D 30.8 Newton. So the charges are going to be assumed to be infinitely small and have no volume and also static in space. The effect of other forces such as that of gravity is also going to be negligible. So we can start by drawing the sort of like a free body diagram or just a diagram of the different forces that's going to be acting on charge M. So first it's going to be uh the repulsive electric force F and on M due to charge N and M being both positive, I'm going to indicate that with this red uh arrow here which is going to be F and on M just like so and we are going to just um take the convention that charge FN on M is going towards the negative direction. Well, charge fo on M is going towards the positive direction because it has an A or when it has an attractive uh force acting because charge M is positive and O is negative. So they're attracting one another. So this is going to be indicated with the blue arrow fo on M and it is going to be going in the convention that we just decided to the positive direction of the axis. So the first step that we're going to take is to use Kla's Law in order for us to calculate the forces acting on charge M. So I'm just gonna write Hulu's Law here. So let's start with FN on M. The law is going to be equals to F equals to K multiplied by in the absolute value of QM or charge of M and QN or a charge of N divided by R squared between M and N. So in this case, K is just a constant which is going to be equal to 9. times 10 to the power of nine new 10 m squared divided by coom squared multiplied that with the absolute value of the M charge, which is 1.3 micro colum. So we want to put that in uh si unit. So it's going to be in Coolum. So we have to multiply 1.3 with 10 to the power of negative six coon. And then multiply that with charge of N which is going to be 2.1 nano Colum. So then this will be 2.1 multiplied by 10 to the power of negative nine cool oops absolute value. And then all of that is going to be divided by the distance between M and N which is going to be two centimeters apart. So that has to also be an si so it has to be 2.0 times 10 to the power of negative 2 m squared. Calculating this, we can get FN on M to then be equal to 6.1 times 10 to the power of negative two Newton. Similarly, we want to do the same thing for charge or 440 on M. So we want to use the same cos law K multiplied by QO multiplied by QM divided by R squared of mo. So substituting the value we have 9.0 times standard of power of nine new 10 m squared divided by colum squared for the K or the constant multiply that with charge of O in absolute value which is going to be 1.6 times 10 to the power of negative six COOM because it is negative 1.6 micro colum multiply that with the charge of M which is 1.3 times 10 to the power of negative six colum divide that with the distance which is going to be three and 3 m. So 3.0 times 10 to the power of negative 2 m squared. And then this will obtain the FO on M to then be equal to 20.8 Newton. So the next step that we want to do is to take the factor sum of the two forces to determine the net force of the forces on charge M. So then F on M Sigma F on M will be equal to FN on M plus F 01 M. And we have to remember that we essentially um by convention decided that FO on M to be positive and FN on M to be negative. So therefore FN on M that we found to be 6. times 10 to the power of negative two. Newton is going to be negative 6.1 times 10 to the power of negative two Newton. I'm gonna put that and prevent the CS plus FO on M which is going to be 20.8 Newton. And calculating this Sigma F on M will then be equal to 20.7 Newton round in rounding it down a little bit. So the final answer for this practice problem is going to be 20. Newton, which will correspond to option B in our answer choices and option B with the net force being 20.7 Newton will be the answer to this particular practice problem. So that'll be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this one. Thank you.

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What is the magnitude of the net force on charge A in FIGURE EX22.17?

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