Static Electricity -07-  Deflection Angle of Charged Particles - video Dailymotion

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Particle A, with a charge q = 0.8 μC, is clamped to a vertical wall. A small sphere B, with a mass of 50 g and the same charge, is suspended from the wall by an insulating thread 25 cm long. The point of suspension is 25 cm above particle A. Determine the angle θ that the thread makes with the wall in equilibrium. Suppose g = 10 m/s².  Watch the video related to *STATIC ELECTRICITY* : https://dailymotion.com/playlist/x80pkv  By following this channel it is very useful for the development of this channel. The Dailymotion application is easy to install on any smartphone, and does not take up much memory space.

Transcript

00:00Hi friends, if you feel tired while studying, don't forget to take a break.

00:07An electrically charged particle is stuck to the wall.

00:12Another electrically charged particle connected to a rope is right next to the particle.

00:18Because the signs of the charges are the same, the particle connected to the rope will be lifted upwards.

00:25Now, we are asked to calculate what is the value of the angle formed between the rope and the wall surface.

00:34Maybe you can't understand this problem properly, just watch this animation.

00:40Here is an electrically charged particle stuck to the wall surface.

00:46Particles like this can't move anywhere.

00:50Another particle connected to the rope is right next to the particle on the wall.

00:56This means that the length of the string is exactly the same as the distance from the axis to the particle on the wall.

01:03To differentiate them, let's call them particle 1 and particle 2.

01:10Since the signs of the charges are the same, the two particles will move away from each other.

01:17Of course, the particle connected to the string will be lifted upward.

01:23The particle will stop moving once the force equilibrium is reached.

01:29At that point, the angle formed by the string and the wall surface is theta.

01:36Before going any further, the length of the string is L, likewise, the distance from the axis of the string to particle 1.

01:46Now let's analyze the forces acting on particle 2.

01:53Particle 2 is around the Earth. The force of gravity will pull particle 2 downward.

02:01The column force is the central force.

02:04This force will be in line with the line connecting particle 1 and particle 2.

02:11The rope is in a taut condition, there is a tension force in the rope according to the direction of the rope.

02:18It turns out that three forces act simultaneously on particle 2.

02:24If you still remember Lamy's theorem in the series of dynamics of motion, this theory requires several angle values.

02:32Because the length of the sides of the triangle is the same, of course it is an isosceles triangle.

02:39In an isosceles triangle, the two angles are equal.

02:45The sum of the angles in each triangle is 180 degrees.

02:51Simply put, beta is equal to 90 minus half theta.

02:58Now extend the string line.

03:01If so, this angle is theta.

03:06The lines of force of the gravitational force will be parallel to the wall surface.

03:12This angle must also be equal to theta.

03:18Overall, angle theta plus angle beta will equal 90 plus half theta.

03:25The sum of the supplementary angles is 180 degrees.

03:30This angle is 180 minus theta.

03:34These are the two angles we need.

03:40Using Lamy's theorem, mg over the sine of 90 plus half theta is equal to f21 over the sine of 180 minus theta.

03:52The tricky part is getting the value of f21.

03:56This is because we need the distance between particle 1 and particle 2.

04:01Let's assume that distance is r.

04:07We can draw a line from the axis to the midpoint of r.

04:11This will form a right angle since it's an isosceles triangle.

04:18Sin half theta is equal to half r over L.

04:24From here, r is equal to 2L sin half theta.

04:30Knowing this value, we can calculate the magnitude of the column force.

04:34F21 is equal to kq1 q2 over r squared.

04:40The magnitudes of the charges are the same.

04:43The magnitude of the column force is kq squared over 4L squared sin squared half theta.

04:50Based on the trigonometric identity, the sine of 90 minus half theta is equal to the cosine of half theta.

04:57And the sine of 180 minus theta is equal to the sine of theta.

05:04The sine of theta itself can be written as 2sine of half theta cosine of half theta.

05:11On both sides there is a factor of cosine of half theta.

05:15Just cross out those two values.

05:20We can write this equation as the sine of half theta to the power of 3.

05:27Just substitute some of the values listed on the problem sheet.

05:33The numbers are very complicated, just use a scientific calculator.

05:40If I'm not mistaken, the sine of half theta is equal to 0.2846.

05:48From here half theta is equal to 16.5 degrees.

05:52Theta is equal to 33 degrees.

05:58Although it looks simple, this problem is really complicated.

06:02Happy learning everyone.

06:04Happy learning everyone.

Static Electricity -07- Deflection Angle of Charged Particles

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