Motion of Charged Particles in Electric Fields

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Throw a cricket ball sideways from a rooftop, and it curves down along a parabola, right? Now fire an electron sideways between two charged plates, where gravity is far too tiny to matter. So does the electron fly in a straight line? Hold that thought; by the end of this video, we will settle it with real numbers. Questions like this come up often in JEE, so let's begin.

0:29A charge in a uniform field

So, put a charge q, of mass m, in a field, E vector; the little arrow on top means a vector. The force is q times E vector, so by Newton's second law, the acceleration vector is q times E vector, over m. Right? Now, suppose the field is uniform, the same size and direction everywhere. Then q, E and m are all fixed. So what kind of motion do we get? Yes, exactly, the acceleration is constant, like a ball under gravity! So the equations of motion work here: v equals u plus a times t, s equals u times t, plus half times a times t squared, and v squared equals u squared, plus two times a times s. Clear? Now fire a positive charge straight against the field with speed v. The force acts against the motion, like gravity on a ball thrown up. With final speed zero and a retardation of q E over m, the third equation gives s equals m times v squared, over two times q times E, okay? And if gravity also acts, add the accelerations as vectors. g effective, the g with the short word eff below it, equals g vector, plus q times E vector over m. Got it?

1:49Between two parallel plates

Now, back to our puzzle. Say a charge enters between parallel plates of length capital L, with speed u along them, and the field, E vector, points across. Which part of the velocity changes, along the plates or across them? Yes, only the part across the plates changes. Along them there is no force, so the velocity along them, v with a small x below it, stays equal to u. So the time inside is capital L over u. With me? Across the plates, the charge starts with zero velocity, and has constant acceleration, like the falling ball. So the deflection y is half times a times t squared. Put in a and t, and y equals q times E times L squared, over two times m times u squared. Okay? Now a neat form. With the kinetic energy, capital K, equal to half times m times u squared, y equals q times E times L squared, over four times capital K, right? As it leaves, the path makes an angle theta, the Greek letter on screen, with the plates. tan theta is v with a small y, the velocity across, over the velocity along, which is a times t over u. So it equals q times E times L, over m times u squared. Now, once the charge leaves the plates, does its path keep curving? It does not, so be careful here! Outside the plates there is no field and no force, so the charge moves in a straight line at that angle. Got it?

3:25A charged pendulum in a horizontal field

Now a pendulum, which comes up often. The bob has mass m and charge q, in a horizontal field. So g effective vector is g vector plus q times E vector over m, and the two are perpendicular, the upside down T sign on screen. Clear? By Pythagoras, g effective is the square root of g squared, plus q E over m, whole squared. The string rests along g effective, at theta from the vertical, with tan theta equal to q times E, over m times g. Okay? Now, the field is horizontal, and the period depends on gravity, which is vertical. So does this field change the period at all? Yes, it does, so careful here! With string length capital L, the period, capital T, is two pi times the square root of L over g effective, pi being the Greek letter. Since g effective is bigger than g, the period gets smaller. With me?

4:23Example 1: electron speed and distance after 1 ns from rest

Let's try our first example. An electron starts from rest in a field of ten to the power three newton per coulomb, written N C to the power minus one. Find its speed and displacement after one nanosecond, n s on screen. Small e over m, the electron's charge over its mass, is one point seven six times ten to the power eleven coulomb per kilogram. Pause the video and try it first. Done? Okay, let's check. The acceleration is small e times capital E, over m, so e over m times the field. That gives one point seven six times ten to the power fourteen metre per second squared. Okay? It starts from rest, so v equals a times t, with t equal to ten to the power minus nine second. That gives one point seven six times ten to the power five metre per second. Right? Now the displacement. Can we just multiply this speed by the time, v times t? We cannot, because the speed grew from zero! So s equals half times a times t squared, with t squared equal to ten to the power minus eighteen. That gives eight point eight times ten to the power minus five metre, or eighty eight micrometres, the mu m on screen. Plain v times t would give double that. Clear?

5:48Example 2: electron deflection and angle between 5 cm plates

Now a harder one. An electron enters midway between two plates with speed two times ten to the power seven metre per second, parallel to them. The plates are five centimetres long, and the field is ten to the power four newton per coulomb. Find its deflection on leaving, and the angle of its path. Pause the video and try it first, with the same e over m. Done? Okay, let's check. The speed along the plates stays the same, so t is capital L over u. That is zero point zero five metre, over two times ten to the power seven, which gives two point five times ten to the power minus nine second. Right? The acceleration across the plates is small e times capital E, over m. That gives one point seven six times ten to the power fifteen metre per second squared. Okay? Now y is half times a times t squared, and t squared is six point two five times ten to the power minus eighteen. Half of one point seven six, times six point two five, gives five point five. With the powers of ten, y is five point five millimetres. With me? Now the angle. The electron moved five point five millimetres sideways over fifty millimetres. So tan theta is just y over capital L, which is zero point one one, right? Oh no, careful, that is the trap! The path is curved, so we need the velocity as it leaves: tan theta is a times t over u. The top is four point four times ten to the power six; divide by two times ten to the power seven, and we get zero point two two. So our puzzle is solved: the electron does bend, just like the cricket ball.

7:38Quick revision: qE/m, plate deflection, charged pendulum

So, quick revision. In a uniform field, q E over m is a constant acceleration; with gravity, add it to g as a vector. Between plates, the velocity along them stays fixed, y and tan theta follow these formulas, and outside, the path is straight. Against the field, a positive charge stops after m times v squared, over two times q times E, and a charged pendulum swings with g effective. Now try the practice problems on this page. All the best!