A +8.75-mC point charge is glued down on a horizontal frictionles...

Question

A +8.75-mC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50-mC point charge by a light, nonconducting 2.50-cm wire. A uniform electric field of magnitude 1.85 * 10^8 N/C is directed parallel to the wire, as shown in Fig. E21.34.

(a) Find the tension in the wire.

Accepted Answer

Hey, everyone today, we're dealing with the problem about horizontal equilibrium. So we're being told that we have two tiny charged fears that are modeled as point charges here that are being connected using a massless non conducting rod. Let's highlight that using a massive non conducting rod of a length of 3.21 centimeters. Now, one sphere has a charge of negative one micro columns, Whereas the other has a charge of 7.2, micro columns, The sphere of charge -5.1 Micro columns has a fixed position while the other is free that somebody is placed in an even external electrical field of magnitude two times 10 to the eight newtons per Coolum. And with this information, we're being asked to find what the tension is in the rod. Now, as I mentioned earlier, this is a problem about horizontal equilibrium. And we know that for horizontal equilibrium there is a oops excuse me. For horizontal equilibrium, we know that the sum of all forces since we're dealing with the horizontal direction here should be zero. This is the postulate for equilibrium. We're also going to need, we utilize columns, law which they said force is equal to excuse me, constant multiplied by two charges, the magnitude of two charges divided by R squared. Now we'll come back to that in just a second. But let's write down what all we know so far, we know that one micro column, one micro Coolum is equal to one times 10 to the negative sixth columns. We also know that Q one is fixed and has a charge of negative 5. micro columns or using the conversion factor that we just used will be five negative 5.1 times 10 to the negative sixth columns. We can utilize similar logic for Q two Which will say is positive 7.21 Times to the -6 Columns. R The distance is noted as 3.21 cm And we can convert this to meters by saying that one m is equal to 10 to the negative or sorry is 10 to 2 centimeters, 10 square centimeters. I should say Because it's an answer of 3.2, 1 times 10 to the -2 m. And the magnitude of the electric field is noted as 2.0 times 10 to the 8th newtons per Coolum. We also know one more thing and that is that force force can also be defined as a charge multiplied by the magnitude of the electric field that's in. So with this in mind, let's go ahead and take a look at our diagram, we know that the charges have opposite opposite signs, which means that the force between them will be attractive and the positive oops and the positive force, excuse me, the positive force will be attracted to the right and it'll be attracted to the right because this uh negative 5.1 micro column charge is fixed, which means the right or the left charge, which is free will gravitate towards the right. So We have noted that Q1 is this right charge. So since Q1 is fixed, Q one is fixed, so the sum of all charges on Q one, Q1 will equal zero because it's fixed. So I hope that makes sense. Anything that is around that point go all to ensure that it is stationary. Now Q two is free and the rod ensures that queue to remain stationary or excuse me, the rod helps make sure that it um remains stationary by pushing or pulling against the fixed uh Q one, right? Because even though it is free, it's still being held at this distance of 3.21 centimeters by this rod, it is providing that tension force that is keeping this charge in place. Even though it's being attracted to the Q one, the negative 5.1 cool in charge. So The Rod will provide that force that pull or push to balance any inequality in the electric forces so that the net force to that net force on Q two Is also equal to 0 to help us maintain or establish equilibrium, right? So let's write this out and I'll use blue, we know that the sum of all forces The X Direction should be equal to zero. And if the positive charge experiences a force in the direction of an electric field, electric field force On the positive charge, Q two will be to the left as it's expressed here. So we can say that if we're summing up the forces, the tension in the rod plus the attractive force, put attractive force minus the force acting against the attractive force, the force of the electric field will be equal to zero, right? So solving for tension or rearranging for tension and let me scroll down a bit. So we just have a wee bit more space equating for tension gives us uh the man or the force due to the electric field minus the force of attraction which we can now substitute in values. We have the 7.21 or 7.7 point 21 times 10 to the negative six cooling charge multiplied by the 2. times 10 To the 8th Newton cool um electric field minus minus The constant nine times 10 to the 9th multiplied by the charge. And remember I didn't write it above, but we have to deal with the absolute value of said charges. So it'll just simply be 5.1 times 10 to the negative six columns multiplied by 7.21 times to the negative six columns divided by divided by R squared, which is simply 3.21 times 10 to the negative two m squared. So solving and equating all of this, we get that tension is equal to 1.12 times to the third Newton's or answer choice C. And what this means is that the rod pulls the charge to the right and using Newton's. So the law, the charge pulls the rod to the left as well. So the rod is under constant tension. Attention of magnitude 1. times 10 to the third Newtons. I hope this helps and I look forward to seeing you all in the next one.

A +8.75-mC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50-mC point charge by a light, nonconducting 2.50-cm wire. A uniform electric field of magnitude 1.85 * 10^8 N/C is directed parallel to the wire, as shown in Fig. E21.34. (a) Find the tension in the wire.

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A +8.75-mC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50-mC point charge by a light, nonconducting 2.50-cm wire. A uniform electric field of magnitude 1.85 * 10^8 N/C is directed parallel to the wire, as shown in Fig. E21.34.
(a) Find the tension in the wire.