2. Two parallel metal plates are charged so that an E field...

2. Two parallel metal plates are charged so that an E field flows from left to right between them. They are separated by a distance of $1 \times 10^{-5}$ meters. A charge of of $-3 \times 10^{-18}$ C and a mass of of $4 \times 10^{-24}$ kg is released from the left plate at an initial speed of 4000 m/s. What is the voltage across the plate necessary to slow this particle to 100 m/s by the time it reaches the right plate?

Transcript

0:00 Hi.

00:01 Here in this given problem there are two parallel metallic plates, the left one and the right one.

00:18 Left one is at 0 volt and the right one is at a potential 7 .0 into 10 to the power 3 or we can say this is 7 ,000 volt.

00:30 Gap between them this is 0 .08 800 meter.

00:37 An ion is being created here in a hole in the left metal plate.

00:44 It is moving, it is passing through the region between the two plates and gaining a speed v as passing through the hole in the second plate.

00:56 In the first part of the problem as the ion produced here is being accelerated as the right plate is at positive potential so its potential is high means this is at a positive potential and the ion is being accelerated towards the positive plate so it must itself should be negative so the ion itself should be negative so the ion itself should be negative and this is the answer for the first part of problem.

02:01 Then in the second part of the problem mass of the iron is given as 2 .70 into 10 to the power this is minus 26 kilogram and charge carried by it this is 3 .20 into 10 dash per minus 19 kula magnitude of the charge.

02:26 To find the speed of the the ion speed gained by the ion as it comes out of the right plate we use energy conservation which says kinetic energy gained by the ion that should be equal to electrostatic potential energy which it gains as passing through the region between the plates so using it we get an expression for the speed gained by the charged particle and that is v is equal to the square root of of twice of q into delta v potential difference divided by mass of the charged ion...

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