An electron with a velocity of 174 ms in the positive y...

An electron with a velocity of 17.4 m/s in the positive y-direction enters a region where there is a uniform electric field of 226 N/C in the positive x-direction. The mass of the electron is 9.109 × 10−31 kg. What is the y-component of the electron’s displacement 3.80 μs after entering the electric-field region if no other forces act on it?

Transcript

00:02 Hi, in the given problem there is a uniform electric field along positive x -axis.

00:13 And a charge moving in a direction of positive y -axis enters into this electric field.

00:24 As you know, electric field is always from positive to negative.

00:28 So this charge particle which is actually an electron, then it will enter.

00:33 Into this electric field it will deflect towards the positivity means positive charge like this so we have to find the distance covered by this charge particle in the electric field means deflection of the chart particle in the electric field along y -axis and along x -axis here this deflection will be delta x along x y -axis delta y so the uniform electric field is given as 200 newton per column speed of electron along positive y -axis is given as let it be y and that is equal to 10 .0 meter per second now this electron will move a uniform motion along y -axis as it will experience the force and hence the acceleration only along x -axis.

02:22 The time given in which we have to find these deflections is p is equal to 2 .40 microsecond.

02:32 So in this one, much time that deflection covered the deflection undergone by this electron along y -axis is delta y is equal to distance equals to speed into time v y into delta t means this is 10 .0 meter per second multiplied by 2 .40 into 10 to minus 6 seconds so this delta y here comes out to be to 24 .0 or we can say this is 2 .40 this is it will become 2 .40 into 10 tisper minus 5 meter so this will be the distance the deflection undergone by the electron along y -axis now for its motion along x -axis there will be no initial component of the velocity along x -axis means v i x is equal to 0 and to find its acceleration along x -axis using the expression for the electrostatic force f equals to the product of charge with the electric field and also it is given as m a x where a x will be the acceleration along x -axis so this acceleration a .x will be equal to e e by m.

04:16 So as v i .x is equal to 0, a .x is equal to this much.

04:20 And time we know delta t is equal to 2 .40 microsecond...

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