55. Two small balls, each of mass ( 5.0 mathrm g ), are...

55. Two small balls, each of mass ( 5.0 mathrm{~g} ), are attached to silk threads ( 50 mathrm{~cm} ) long, which are in turn tied to the same point on the ceiling, as shown below. When the balls are given the same charge ( Q ), the threads hang at ( 5.0^{circ} ) to the vertical, as shown below. What is the magnitude of ( Q ? ) What are the signs of the two charges? 56. Point charges ( Q_{1}=2.0 mu mathrm{C} ) and ( Q_{2}=4.0 mu mathrm{C} ) are located at ( overrightarrow{mathbf{r}}_{1}=(4.0 hat{mathbf{i}}-2.0 hat{mathbf{j}}+5.0 hat{mathbf{k}}) mathrm{m} ) and ( overrightarrow{mathbf{r}}_{2}=(8.0 hat{mathbf{i}}+5.0 hat{mathbf{j}}-9.0 hat{mathbf{k}}) mathrm{m} . ) What is the force of ( Q_{2} ) on ( Q_{1} ? ) 62. The charges ( q_{1}=2.0 imes 10^{-7} mathrm{C}, q_{2}=-4.0 imes 10^{-7} mathrm{C} ), and ( q_{3}=-1.0 imes 10^{-7} mathrm{C} ) are placed at the corners of the triangle shown below. What is the force on ( q_{1} ) ? 83. The charge per unit length on the thin rod shown below is ( lambda . ) What is the electric field at the point ( P ? ) (Hint: Solve this problem by first considering the electric field ( d overrightarrow{mathbf{E}} ) at ( P ) due to a small segment ( d x ) of the rod, which contains charge ( d q=lambda d x ). Then find the net field by integrating ( d overrightarrow{mathbf{E}} ) over the length of the rod.) 90. A rod bent into the arc of a circle subtends an angle ( 2 heta ) at the center ( P ) of the circle (see below). If the rod is charged uniformly with a total charge ( Q ), what is the electric field at ( P ? )

Transcript

00:01 So here we have two small balls and they are hanging like this and this is the ceiling here.

00:15 So each of these balls they have mass of 5 grams and the length of the string to which it is attached, that length is 0 .5 meters.

00:29 And the angle that these balls make with respect to horizontal, that is five degrees.

00:41 So at this position, the two balls, they are remaining in equilibrium.

00:50 So here, these balls are given same charge, and we need to figure out the magnitude of this charge q contained in the ball so that the balls will remain in this position.

01:03 So here there will be tension.

01:07 If we see the free body diagram of either ball, there will be tension.

01:12 There will be the force mg in the downward direction.

01:17 And as these are having similar charge, there will be repulsive force and that will be directed in this direction.

01:23 So let's see that repulsive forces.

01:28 So here we break this tension into two components.

01:31 So we get t cause 5 degree.

01:34 And here we have t sign 5.

01:36 Degree so k cost 5 degree will be equal to m g and t sign 5 degree will be equal to electrostatic force so we divide it and dividing these equations we get 10 5 degree times mg is equal to f so putting the value of m as 5 divided by 1 ,000 because we'll use the si unit and multiplying that by d which is 9 .81 we get 0 .049 and then we multiply with tan of 5 degree to get the force f so here tan of 5 degree times 0 .049 if we multiply we get the force as 0 .0043 neutins now we can use the expression of force using culems law.

02:43 So we use k as the charges are same.

02:47 So q square by separation square.

02:50 So separation we can figure out by considering this right angle triangle, we can get this side of this right angle triangle as 0 .5 times sine 5 degree.

03:04 So the total separation will be this times 2.

03:08 So it will be sine 5 and we will square this because it is r square and this is equal to 0 .043.

03:18 So from here we get the value of q square as 0 .0043 times sine square 5 degree divided by the value of k that is 9 into 10 to power.

03:35 So if we check the value of sine square 5, this comes out to be .0076.

03:48 This will be .0076.

03:52 So from here we can find out the charge q.

03:57 So if we multiply the numerator again .0076 times .0043, and then we divide it by 9.

04:09 We get 3 .63 times 10 to par minus 15 that's equal to q square so we can write this as 36 .3 into 10 raise 2 minus 14 so we just take one decimal to the right or basically this can be understood by just multiply and dividing by 10 so we get sorry here we get minus 60 so now we can take the square root to get the charge so that will be approximately 6 into 10 to power minus 8 so we can write this as 0 .6 into 10 to power minus 7 it will become or we can shift it to the right and we can write it as 60 into 10 to power minus 9 so that will be 60 nanoculum that will be the charge contained in each box here now similarly the next one here we have the two point charges which are having values two microculum so 2 into 10 to power minus 6 column another charge for microculum so their locations are given so this position vector is for this charge 1 is 4 i minus 2 j plus 5k and position vector for another charge is 8 i plus 5 j minus 9k so we have to figure out the force on q2 by 1 so basically force on charge 2 by charge 1 so that will be given by k k k u1 q2 2 by r square also this force is a vector so this will be in the direction r2 .1.

06:30 So we'll figure out that later on.

06:32 But first let us find out r21 vector.

06:37 So that will be r2 vector minus r1 vector.

06:40 So that will be 4i plus 7j minus 14k.

06:47 This is r21 vector.

06:50 So if you find the magnitude, that will be r and it will be root of 4 square plus 7 square plus 14 square.

06:59 So when we add them all, we're going to get here 16 .2 approximately.

07:10 So r will be 16 .2.

07:12 So the force here that we will get, we can write, we can improve this.

07:18 We can make this r cube and we can make this as r to one vector.

07:27 So k is 9 into 10 to par 9.

07:29 The two charges, their product is 8 into 10 to power minus 12.

07:33 And r cube if we do that will be 4 .2 4 .2 times 10 to par 10 to par 3 and r21 vector that we will put here just this result so 4 i plus 7j minus 14 k this will be the complete force and when we simplify this we're going to get the required force here as so this will become 10 to power 6 and this will be 10 to power minus 6 so force will be in micronutons and now we just simplify it further approximately we get 17 .1 times 4i plus 7 j minus 14 k micronutons so this is the required force so here we have another problem.

08:54 So there are three charges that are placed at the corners of this right angle triangle.

09:00 So this is q1.

09:01 Here we have q3...

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