HW I 1. An infinite plate is uniformly charged with charge...

HW I 1. An infinite plate is uniformly charged with charge density \sigma. Normal vector to the plate is \hat{i}. The plate oscillates with velocity is \(v(t) = \theta(t)v_0 \cos(\omega t)\hat{k}\). a. Find magnetic and electrical field as a function of time (the fields satisfy the radiation condition at both infinities \(x = \pm \infty\)). Ignore the static electrical field. b. What is the radiated power (per unit of surface area)? This is some time called intensity of the radiation [W/m²] c. What is the external force applied to the plate to maintain its oscillation (force per surface of unit area)

Transcript

00:01 Hello everyone.

00:02 So in the question it is given that we have two parallel plates is shown in this figure.

00:08 Okay.

00:09 So this parallel plate is having a positive surface charge density and this is having a negative surface charge density.

00:18 Okay.

00:18 The positive plate is at 60 volts.

00:21 Okay.

00:21 And the negative is at zero volt.

00:24 Now, these plates are square plates having a length of 8 centimeter.

00:30 Okay.

00:31 And these plates are 1 centimeter apart.

00:35 Now we have to calculate the magnitude of electric field between the plates and we also have to indicate the direction.

00:43 Okay? so the formula to calculate the electric field is given as e is equals to minus v divided by d.

00:58 Okay? so now this voltage is 60 divided by the distance between them is 1 cm so converting it into meters okay so it will be 0 .01 meters okay which gives us electric field as 6 ,000 volt per meter okay so we got the electric field magnitude as 6 ,000 volt per meter okay so we got the electric field magnitude as 6 ,000 volt per meter okay so this is the answer of the the first question now it is given that an electron is allowed to pass between the parallel plates okay so as it is having a negative charge so it will get deflected towards the positive charge it will get attracted towards the positive plate okay having a surface charge density positive we have to find what is the force that is acting on the electron okay so the form for calculating the force is given as q multiplied by e.

02:19 Okay? so here q is 1 .6 multiplied by 10 to the power minus 19 okay and electric field is 6 ,000 volt per meter okay so we got the force acting on the electron as 9 .6 multiplied by 10 to the the power minus 16 newton.

02:48 Okay.

02:49 So this is the four setting on the particle.

02:51 Now we have to find if an electron is launched into the field with a speed in the x direction as 4 multiplied by 10 to the power 7 meter per second, then what is the time for the electron to travel from point a to point b? so now over here we have been given the velocity, okay? so velocity in x direction is given as 4 multiplied by 10 to the power 7 meter per second.

03:26 Okay.

03:27 So now we have to find the time.

03:30 So first of all we need to find acceleration.

03:33 So we know that force is equals to mass multiplied by acceleration.

03:38 So from this we got acceleration as force divided by mass.

03:44 Okay.

03:45 So force is 9 .6.

03:47 Multiplied by 10 to the power minus 16 and masses 9 .1 multiplied by 10 to the power minus 31...

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