Electric Flux

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take a flat round plate, and a bowl shaped like a hemisphere, with exactly the same rim. Hold both in the same uniform electric field, facing it. The curved bowl has twice the area of the flat plate. So do twice as many field lines pass through the bowl? Hmm, hold that thought. The idea we need is electric flux, and questions on it come up often in JEE, so let's begin.

0:32Flux through a flat area

So, before flux, we need one new idea, the area vector. We write it as capital A with an arrow on top, and the arrow means a vector, okay? Its size is the area, capital A, and its direction is along the normal, the line perpendicular to the surface. Clear? Now, electric flux. We write it as phi, the Greek letter on screen. For a flat area in a uniform field, phi equals E vector dot A vector, the dot product. So phi equals E times A times cos theta, where theta, another Greek letter, is the angle between the field and the area vector. Right? Now let me ask you this. Suppose you slowly turn the surface in the field. When is the flux the biggest, and when does it become zero? Yes, exactly! When the surface faces the field, the normal lies along the field, so theta is zero degrees, the small circle meaning degrees. Cos zero is one, so phi equals E times A, the biggest value. Turn it edge-on, and theta is ninety degrees, so the flux is zero. The field just slides along the surface. Got it? Now, here is a common trap. Suppose a question gives you the angle between the field and the surface itself. Can you put that angle straight into cos theta? You cannot, and that is the trap! Theta is always measured from the normal, not from the surface. So when the angle with the surface is given, theta is ninety degrees minus that angle. With me? Now, when the field is not uniform, or the surface is curved, break the surface into tiny pieces, each one d A vector, and add up E dot d A for all of them. That sum is an integral, the long S sign on screen. So phi equals the integral of E vector dot d A vector. Okay? Finally, the unit. Newton per coulomb times square metre gives newton metre squared per coulomb, written as capital N, m squared, capital C to the power minus one. And since a newton per coulomb is a volt per metre, this is also volt metre, the V m on screen.

2:39Closed surfaces: the normal points outwards

Clear? Now, closed surfaces, like a balloon or a box, which completely enclose some space. For these, the normal always points outwards. Right? So think about a field line that enters the surface. The field points inwards there, but the normal points outwards. So is that flux positive or negative? Yes, it is negative, exactly! The angle with the outward normal is more than ninety degrees, so cos theta is negative. So lines leaving give positive flux, phi greater than zero, and lines entering give negative flux, phi less than zero. Got it? Now put any closed surface, say a box, in a uniform field. Some lines enter it, and some leave it. So what is the net flux through the whole box? It is zero, yes! In a uniform field, every line that enters the box also leaves it, so the negative flux going in cancels the positive flux coming out, and the net flux is zero. Remember, this needs a uniform field. Okay?

3:44Topper's trick: use the flat rim

Now a topper's trick, built on that zero. In a uniform field, the flux through any open surface has the same size as the flux through its flat rim, the flat area that the rim encloses. Here is why. Close the open surface with that flat area, and the net flux is zero, so whatever enters through one part leaves through the other. With outward normals the two signs are opposite, but the sizes are equal. Clear? So, back to our puzzle, the bowl and the flat plate with the same rim. The bowl has twice the area. So how much flux passes through the bowl? The same as the plate, exactly! Take a hemisphere of radius capital R, with its axis along the field. Its rim encloses a circle of area pi R squared, so the modulus of phi through the curved face is pi R squared times E, okay? The two vertical bars mean modulus, the size without the sign, and the small word curved below phi names the face. That solves our puzzle!

4:44Example 1: flux of a 3i + 4j field through a yz-plane area

Let's try our first example. The field, E vector, is three i hat plus four j hat, in brackets, times ten to the power three newton per coulomb. Here i hat and j hat, the letters with little caps, are unit vectors along x and y. Find the flux through an area of zero point two square metre, lying in the y z plane. Pause the video and try it first. Done? Okay, let's check. Three, four and five make a right triangle, so the size of the field is five times ten to the power three. Then the flux is five thousand times zero point two, which is one thousand, right? Oh no, careful, that is the trap! Flux needs the normal to the surface, and any surface in the y z plane has its normal along the x axis. So the area vector is zero point two i hat square metre. Okay? Now take the dot product, part by part. Three times ten to the power three, times zero point two, plus four times ten to the power three, times zero. The j part of the field runs along the surface, so it gives no flux at all. With me? So phi equals six hundred newton metre squared per coulomb. The answer is six hundred, not the one thousand from the trap. Always find the normal first. Clear?

6:04Example 2: flat, curved and net flux of a closed hemisphere

Now a harder one. Put a closed hemisphere of radius capital R, the bowl together with its flat lid, in a uniform field E, parallel to its axis. Find the flux through the flat face, through the curved face, and through the whole surface. Pause the video and try it first. Done? Okay, let's check. Say the field lines enter through the flat face. The field is uniform, so every line entering the flat face leaves through the curved face. Right? Now, the flat face is a disc of area pi R squared, facing the field. So is its flux plus pi R squared times E, or minus? It is minus, yes! The outward normal of the flat face points against the field, so theta is one hundred and eighty degrees, and cos theta is minus one. The lines enter here, so phi flat equals minus pi R squared times E, okay? And the same lines leave through the curved face. So phi curved equals plus pi R squared times E, just as our rim trick said. Got it? Now add the two for the whole surface. Minus pi R squared times E, plus pi R squared times E, gives zero, because a closed surface in a uniform field always has zero net flux. So the bowl in our puzzle gets exactly as much flux as the flat plate. Clear?

7:30Quick revision: EA cos theta, closed surfaces, rim trick

So, quick revision. For a flat area in a uniform field, phi equals E times A times cos theta, with theta from the normal. For a closed surface, the normal points outwards, leaving gives plus, entering gives minus, and a uniform field gives zero net flux. And an open surface in a uniform field has the same size of flux as its flat rim. Now go and try the practice problems on this page. All the best!