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.
(b) What would the tension be if both charges were negative?

(b) What would the tension be if both charges were negative?

Accepted Answer

Everyone today, we're dealing with the problem regarding columns law. So we're being told that we have two balls that are charged to plus 5.1 micro columns and plus 7.21 micro columns that are used in an electrostatic experiment. Now, they're joined together by a massless non conducting Rod with the length of 3. cm. And the 5.1 Micro Column Charge Ball which will label Q1 is in a fixed position while the other charge ball is free. The assembly lies within an electric field and even electric field with a magnitude of two point oh times 10 to the eighth newtons per column. With this, we're being asked to determine the tension or compression in the connecting rod. So we're given the diagram below. But before that, let's just write out what we need to use. So one of the key or two key formulas that we're going to need our columns law. They said columns, constant multiplied by the absolute value of two charges multiplied together, divided by R squared will tell you the force acting upon them. We can also recall that F force can also be equivalent to a force multiplied or a charge multiplied by the magnitude of the electric field that is in with this. Let's also write down what we know. So far Q one is plus 5.1 micro columns. And if we recall one micro column is equal to 10 to the negative sixth columns, so we can rewrite this as 5.1 times 10 to the negative six columns. Similarly, Q two is positive 7.21 times 10 to the negative sixth columns Are, in this case, is the length of the Rod. So it's 3.2, 1 cm, but we want this in meters. So we multiplied by a conversion factor. So recall one m is equal to 10 to the second cm. So this gives us 3.2, 1 times 10 to the negative 2nd m. And the magnitude of the electric field can be written as 2.0 times 10 to the 8th newtons per column. That's what we're given. So we know that Q one And let me use my pointer for this, we know that Q1, Q one, the plus 5.1 charge is fixed, right. So if we take equilibrium, the forces act because the rod is sustaining the two charges in place if we consider equilibrium, right, which they set the sum of all the forces in the x direction are equal to zero. While this means that the sum of all the forces acting upon Q one must also be equal to zero. Furthermore, it's also fixed, which means it cannot move. So which means there are the net force acting upon it will be zero. Similarly, Q two is free. However, the rod is keeping this rod is keeping it in place, which means the sum of all the forces on cute too Must also be zero. There's something the net force that is acting upon it. Now, one thing we need to take note of is that positive charges experience of force in the direction of an electric field, right? So these are two positive charges. So already since they're both positive will experience a repulsive force in either way, they repulse each other. But since Q one is fixed Q two will move away, right? But it's being held in place by the rod, we also know that again, positive charges the experience of forcing the direction of an electric field. So the electric field has noted here is pointing towards the left since it's pointing towards the left, that means that the force that is experienced by you too, by the electric field is also towards the left. So let's think about this, the sum of all forces. Let me write that a little better, the sum of all forces in the X direction Is equal to zero. Well, if we take the positive direction to be towards the right and the negative direction to be towards the left while the tension or whatever we're looking for, minus the repulsive force, f repulsive reports minus the electric force because it's also moving this way. So we have the repulsive force, this is F R as well as F due to E are going this way. But the tension is pulling this way because that's what's keeping the charge from shooting off completely into the other direction, her pulse and this is F of the magnetic field. So this sum will be zero. So if we're looking for the tension, well, before we do that, let's rewrite these, we can write this as cool um Law Q one cute too over R squared -62 E Because we're not concerned. Oops, we're not concerned with the movement of uh Q1 because it's fixed will be zero. So the net force acting on Q1 is already zero. So rearranging for T we then get T is equal to okay. And Q one cute too over R squared plus Q two E and substituting in our values, We get columns constant 9.0 times 10 to the ninth multiplied by Q one which is 5.1 times 10 to the native 6th column multiplied by 7.21 times 10 to the negative six school. Um Oops, let's right that let's move this over just a wee bit. So I have a little more space to write 10 to the 96th. Cool um Over uh It'll be 3.2, 1 times 10 to the night of second squared plus 7.21 times 10 to the sixth columns, multiplied by the magnitude of the electric field is 2.0 times 10 to the eighth per Coolum, which will give us a final answer of 1.7, 6 times 10 to the to the third, excuse me to the third newtons. So the tension in the connecting rod is therefore, excuse me, the tension, the connecting rod is therefore answer choice, see 1.7, 6 times 10 to the 3rd Newtons. And this is significant because the forces are repelling each other, which means there will be tension created, it won't be compressing the rod together. I hope this helps. And I look forward to seeing you all in the next one.

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