In the figure, a uniform, upward electric field E of...

In the figure, a uniform, upward electric field E of magnitude 2.00×10^3 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 4 cm and separation d = 2.00 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v_0 of the electron makes an angle ? = 45° with the lower plate and has a magnitude of 5.64×10^6 m/s. Will the electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not, enter the vertical position at which the particle leaves the space between the plates. The next electron has an initial velocity which has the same angle ? = 45° with the lower plate and has a magnitude of 4.45×10^6 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not, enter the vertical position at which the particle leaves the space between the plates.

Transcript

00:01 Hello, the question is taken from physics and the question is, in the figure, a uniform upward electric field of magnitude, 2 into 10 to the power 3 nanonuton per column has been set up between two horizontal plate.

00:18 By charging the lower plate positively and the upper plate negatively.

00:24 The plate has a length of 4 cm and the suppression is 2 cm.

00:28 An electron is then short between the plates from the left edge of the lower plate.

00:35 Initial velocity v0 of an electron makes an angle theta is equal to 45 degree with the lower plate and has a magnitude of 5 .64 into 10 to the power 6 meter per second.

00:47 Will the electron strike one of the plates? if so, what is the horizontal distance from the left edge? it is the horizontal distance from the left edge okay if not enter the vertical position at which the particle leaves the space between the plate so let us evaluate the distance or by using the equation of motion okay e is given to us that is 2 into 10 to the power 3 newton per column acceleration is corresponding to this electrical which is q e by m so q e is 1 .6 into 10 to the power my 19 e is 2 into 10 to the power 3 mass of electron is 9 .1 into 10 to the power minus 31 so this will give you the value of acceleration is so let me evaluate this is equal to 1 .6 e minus 19 multiply 2 e 3 and divided by 9 .1e minus 31 okay so that is 3 .52 into 10 to the power 14 meter per second square which is the required value of acceleration now initial velocity v0 course of theta component is equal to 5 .46 into 10 to the power 6 theta is i think 45 degree divided by square root 2 meter per second in the similar way vertical component is v0 sign of theta that is equal to 5 .46 into 10 to the power 6 divided by square two meter per second since the direction of electric field is from bottom to upward also upward plate is negative okay so electron will move in this path due to repulsion okay so it will move from here so we need to evaluate the horizontal distance at which it strike the plate so everything is given to us so let me evaluate the horizontal distance that it travels or the initial position is also given to us okay so let me evaluate first the maximum height we know that h max is equal to u square sine square theta over 2a so that is v0 square sine square theta divided by two acceleration so that is 5 .46 into 10 to the power 6 whole square divided by 2 into acceleration that we have already valued that is 3 .52 into 10 to the power 14.

03:50 Okay, so maximum height that up to which the particle will go is equal to 5 .46e6 whole square divided by 2 divided by 3 .52 e 14.

04:07 So that will be 0 .0 423 meter.

04:12 So let's.

04:12 Let me check what is the value of d d is equal to two centimeter so from here we can write we can write it as 0 .02 meter so it's strike the plate at the height 0 .0 4 2 meter okay so now we need to evaluate at what value of d it strike the plate so at what value of a we can say horizontal distance it will strike the plate okay so we know that x is equal to to v course of theta or initial course of theta into t so the time it will take to reach at this height is equal to we know that x y is equal to or the horizontal distance is equal to y initial is 0 so uy into t plus half a t square and the direction of acceleration is also in the downward so it will be negative and velocity is in the upper direction so d is equal to 0 .02 is equal to uy is uy so we have already that is v sign of theta that is 5 .4 6 into 10 to the power 6 by square 2 into t minus acceleration is 3 .52 divided by 2 into 10 to the power 14 into t square.

05:44 So let me solve this quadratic equation...

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