INT The surface charge density on an infinite charged plane is −2...

Question

INT The surface charge density on an infinite charged plane is −2.0×10^−6 C/m^2 . A proton is shot straight away from the plane at 2.0×10^6 m/s . How far does the proton travel before reaching its turning point?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, a flat infinitely large plane with a surface charge density of negative 2.9 multiplied by 10 to the power of negative six columns per meter squared attracts an alpha particle propelled away from it at an initial velocity of 1.6 multiplied by 10 to the power of 6 m per second. How far does the particle travel before being drawn back towards the plane? OK. So that is our end goal to determine how far the particle travels before being drawn back towards the plane. OK. So we're given some multiple choice answers here. They're all in the same units of meters. Let's read them to see what our final answer might be. A is 4.02 B is 0.08 and C is 3.04 and D is 1.05. OK. So first off, we need to recall and use the work energy theorem when a particle comes to rest the initial kinetic energy of the particle equals the work done by the electric force over distance delta X. So we can write the following and let's call it equation one, one half multiplied by M multiplied by the initial velocity squared is equal to F multiplied by delta X. OK. So recall that the electric force capital F on a charged particle due to any electric field represented by capital E is given by and let's call it equation two, F is equal to the charge multiplied by the electric field. OK. So now we can recall that the electric field capital E created by a uniformly charged surface with surface density, which let's note that surface density is denoted by A. So this is surface density and the electric field, let's call it equation three is equal to a, the surface density divided by E zero E subscript zero. OK. So now we can plug equation three into equation two. And when we do that, we should get equation for which states that F is equal to Q, the charge multiplied by ada divided by E zero. Now we can substitute equation one into equation four. So we can say equation five is one half multiplied by M multiplied by V I squared V subscript I squared. The initial velocity is equal to the charge multiplied by a divided by E zero multiplied by delta X. So note that we now need to sol for delta X So let's rearrange to isolate and sol for delta X. So delta X is equal to M multiplied by V I squared multiplied by E zero, divided by two, multiplied by the charge multiplied by the surface density. So at this stage, we could plug in all of our known variables to sol for delta X. So let's do that. So M represents the mass and we know that the mass of an alpha particle because that's what we are given in the pro is an alpha particle. So when we look it up, we do find out that the mass of an alpha particle is 6.644 24, multiplied by 10 to the power of negative kilograms multiplied by the initial velocity which is given to us in the problem as 1.6 multiplied by 10 to the power of 6 m per second multiplied by E zero, which that constant is eight point multiplied by 10 to the power of negative squared per newtons multiplied by meter squared. OK. And don't forget the squared on velocity. OK. Divided by two multiplied by the charge which was given to us as three point two multiplied by 10 to the power of negative 19 cool lums multiplied by the surface density which is 2.9 multiplied by 10th of the power of negative six S per meter squared. OK. So when we plug that into a calculator, we should get that delta X is equal to 0. m, which is our final answer. So note that the displacement delta X is a distance and also a scalar quantity, meaning that it is always a positive quantity. Therefore, we consider the magnitude of the calculated value will, will result in a positive distance of 0. m. OK. So going back to look at our multiple choice answers, that means that our correct answer has to be the letter B 0.08 m. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Um.

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INT The surface charge density on an infinite charged plane is −2.0×10^−6 C/m^2 . A proton is shot straight away from the plane at 2.0×10^6 m/s . How far does the proton travel before reaching its turning point?

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