Electric Potential

Transcript

0:00Introduction

Hello everyone! Here is a puzzle. Put a charge plus q and a charge minus q a little apart, and look at the point exactly midway. The potential there is zero. So a charge placed there feels no force, right? Hmm, hold that thought. Electric potential and its links with the field come up often in JEE questions, so let's begin.

0:25Electric potential: work per unit charge

So, what is electric potential? We write capital V, and the small capital P below it names the point, okay? V P is the work done by an external agent, per unit charge. Right? Work done in bringing a small test charge slowly from infinity, the sideways eight on screen, to the point P, okay? For one point charge q at a distance r, V equals k q over r, with k equal to one over four pi epsilon naught; epsilon naught, the Greek letter with a small zero below it, is the permittivity of free space. Now, for several charges we added fields as vectors, remember? So do potentials need directions too? No, and that is the beauty of potential! Potential is a scalar, a number with a sign. So V equals V one plus V two plus V three; the small numbers are labels, and the dots mean more terms. Just keep each sign. Got it?

1:24From the field to the potential

Now, the first link. V B minus V A equals minus the integral from A to B of E vector dot d l vector. The long S is the integral sign, the arrows mean vectors, and d l is a tiny step along the path, okay? Does the answer depend on which path we take? No, it does not! The electrostatic field is conservative, so every path gives the same answer. The integral with a small circle is over a closed loop, where E dot d l adds up to zero. With me? In a uniform field, it is simpler. V A minus V B equals E vector dot the bracket, r B vector minus r A vector, the displacement from A to B, okay? Watch the order on the left. Clear?

2:14From the potential to the field

Now the second link, the other way round. E vector equals minus nabla V, where nabla, the upside down triangle, acting on V gives this bracket, and each curly d is a partial derivative along x, y or z. And i hat, j hat, k hat, with little caps, are unit vectors along the axes. Okay? In one dimension, E equals minus d V by d x, minus the slope of the graph of V against x. Now, suppose the field points along plus x, so E is positive, and you walk along it. Does V go up, or go down? It goes down, yes! Positive field means negative slope, so V falls along the field. Right?

2:59Equipotential surfaces

Now let's see this on a picture. Join all the points with the same potential, and you get an equipotential surface, okay? So how much work is done in moving a charge along one? It is zero, exactly, because the potential does not change along the surface. With me? So every small step along the surface has E dot d l equal to zero, and the field is everywhere perpendicular to it. In panel a, the dashed circles cut the field lines at right angles. Clear? So for a point charge, the equipotentials are spheres, the dashed circles in panel a. In a uniform field they are parallel planes, as in panel b, where V one is greater than V two, which is greater than V three. Got it? Finally, where the field is strong, the equipotentials crowd together, since E equals delta V over delta r; capital delta, the Greek triangle, means a small change. Okay?

3:56Trap: V=0 does not mean E=0

Now, back to our puzzle. The midpoint is at the same distance r from both charges, so V equals k q over r, minus k q over r, which is zero. So is the field zero too? No, it is not, and that is the trap! The field of plus q points away from it, towards minus q, and the field of minus q also points towards minus q. So they add, and E is not equal to zero, the equals sign with a slash; a charge there does feel a force. Right? The reverse is also true. Panel c plots V against r. Inside a shell of radius capital R carrying charge capital Q, the field is zero, but V stays at k capital Q over capital R, the flat line marked shell; the curve marked solid is for a uniform solid sphere. Okay? So remember, E is minus the slope of V, not its value. Clear?

4:54Example 1: VA minus VB in a uniform 100 V/m field

Our first example has a uniform field of one hundred volt per metre along plus x; the unit V m to the power minus one is read volt per metre. Find V A minus V B for point A at one comma two and point B at four comma six, in metres. Pause the video and try it first. Done? Okay, let's check. From A to B, x goes up by three and y by four, so the distance A B is five metres. So one hundred times five gives five hundred volt, right? Oh no, careful, that is the trap! Write the displacement first. r B vector minus r A vector is four minus one, times i hat, plus six minus two, times j hat, which is three i hat plus four j hat metres. Okay? The field is one hundred i hat, so the dot product keeps only the x part, one hundred times three. The four metres along y change nothing. With me? So V A minus V B is three hundred volt; the capital V after the number is the unit, volt. Only the displacement along the field matters. It is positive, since V falls along the field. Clear?

6:12Example 2: points where V = 0 for +q and -2q on the x axis

Now a harder one. Charges plus q and minus two q sit at x equals zero and x equals three metres. Find all the points on the x axis where the potential is zero. Pause the video and try it first. Done? Okay, let's check. The distances are mod x from plus q, and the modulus of x minus three from minus two q; the vertical bars mean modulus, a distance without sign. So V is zero when k q over the modulus of x equals two k q over the modulus of x minus three; the double arrow shows it after cancelling k q. Okay? First, between the charges, where zero is less than x, and x is less than three. The distances are x and three minus x, so three minus x equals two x, and x equals one metre. Clear? So the answer is x equals one metre, and we are done, right? Oh no, careful, we are not done! Stopping at the first point is the JEE trap. To the left, x less than zero, the distances are minus x and three minus x, so three minus x equals minus two x, and x equals minus three metres. With me? And to the right, x greater than three? There, x minus three equals two x gives x equals minus three, not in this region. We are closer to the bigger charge there, so there is no zero. Got it? So V is zero at x equals one metre, and at x equals minus three metres. Potential is a scalar, so no directions were needed. Clear?

7:54Quick revision: V = kq/r, E = -dV/dx, equipotentials

So, quick revision. V equals k q over r, added as scalars with signs. V B minus V A is minus the integral of E dot d l, E is minus d V by d x, and V falls along the field. Equipotentials are perpendicular to the field, and V equals zero does not mean E equals zero. You did really well today. Now try the practice problems on this page. All the best!