Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Give a hollow metal sphere some charge. Inside it, the electric field is zero, everywhere. So the potential inside is zero too, right? Hold on to that answer. Today we work through the six concept checks of this unit's problem bank, starting with this one. Ideas like these are often tested in JEE in statement form. So, let's begin.

0:29Check 1: zero field, but not zero potential?

So, check one, our puzzle from the start. Take a thin conducting shell, a hollow metal sphere, and put some charge on it. Inside it, E, the electric field, is zero. Okay? So here is the check. Inside this shell, is V, the potential, zero too? No, it is not zero at all. Look at panel c on the right: V against r, the distance from the centre. The red line marked shell is flat, well above zero; the blue curve inside is for a uniformly charged solid sphere, an insulator, and outside, both follow the same falling curve. Inside the shell, V equals k times capital Q, over capital R, at every point, where capital Q is the charge on the shell and capital R is its radius, marked on the graph. Here k is one over four pi epsilon naught, and epsilon naught, the Greek epsilon with a small zero, is the permittivity of free space. Right? Here is why the field is still zero. The field is minus the slope of the V graph. Inside the shell the graph is flat, so its slope is zero, and E is zero, even though V is not. The field depends on how V changes, not on its value. Clear? And it works the other way too. At the midpoint of plus q and minus q, the potential is zero, but the field is not; the equals sign with a slash through it means not equal. So zero field and zero potential are separate questions. Got it?

2:00Check 2: why no closed field lines?

Now check two. Electrostatic field lines never form closed loops. Can you tell me why they cannot? Let's see. Suppose a field line did close on itself, and we walk along it in small steps, d l, with a little arrow on top. The field E, also with an arrow, points along the line, the way we walk. So E dot d l, the dot product, is positive at every step; and the sign after it means greater than zero. Okay? Now add up all the steps, right round the loop. The integral sign with a small circle on it means the sum around a closed path. So the closed integral of E dot d l would be positive. With me? But the electrostatic field is conservative: around any closed path, the integral of E dot d l is exactly zero. Positive and zero at the same time is impossible, so a closed electrostatic field line can never exist. Clear?

2:57Check 3: the potential in an empty cavity

Now check three. First, recall that in a conductor in electrostatic equilibrium, E is zero inside the metal. So the whole conductor is an equipotential: V has one single value all through it. Okay? Now take an empty cavity inside a charged conductor. Is the potential in the cavity the same as that of the conductor, or different? Let's work it out. An empty cavity has no field inside it, and no charge on its wall. This is called shielding: the metal guards the cavity. Right? And with zero field everywhere in the cavity, V cannot change as you cross it, right up to the walls. The walls belong to the conductor, so V cavity, the V with a small subscript cavity, equals V conductor. So yes, it is exactly the same. Got it?

3:49Check 4: Gauss's law without symmetry?

Now check four, Gauss's law. Take a closed surface. The closed integral of E dot d A over it equals q in, over epsilon naught. Here d A, with its arrow, is a tiny piece of the surface, and q in, the q with a small subscript in, is the charge inside the surface. The left side is the electric flux through the surface. With me? Now take a strange, lumpy closed surface, with no symmetry at all. Is Gauss's law still true for it? Yes, it is always true, for every closed surface, whatever its shape. The flux depends only on the charge inside. Okay? But be careful with one thing. The field E in the integral is due to all the charges, inside and outside. Charges outside change the field on the surface, but not the total flux. Right? So symmetry is needed only when we want to use the law to find the field itself. With no symmetry, the law is still true; it just does not give us E easily.

4:50Example 1: does capacitance depend on the charge on it?

Clear? Now, Example one, the easier one. Does the capacitance of a capacitor depend on the charge on it? Pause the video and try it first. Done? Okay, let's check. Capacitance, capital C, is Q over V: the charge on the plates, over the potential difference between them. It is measured in farad. Okay? Now look at this formula. Put more charge on the plates, and Q goes up, so C goes up too, correct? No, wait, that is the trap, because V does not stay the same. Q and V change together: double the charge, and the potential difference doubles too. So Q over V stays the same. Right? Look at panel a: two parallel plates, a distance d apart, one charged plus and one minus. If each plate has area capital A, the capacitance is epsilon naught times capital A, over d, the gap. No Q and no V in it, only the plates and the gap. With me? And if a dielectric, with dielectric constant capital K, fills the gap, C becomes K times C; the arrow on screen means becomes. So the medium between the plates matters too. Clear? So the answer is no. Capacitance depends only on the geometry and on the medium, never on the charge it carries. That is a classic statement question, so remember it.

6:13Example 2: slab into isolated capacitor, does energy rise?

Now Example two, a little harder. Take a charged capacitor that is isolated, with no battery connected, and push a dielectric slab into it. Does the stored energy increase? The hint: ask what stays fixed. Pause the video and try it first. Done? Okay, let's check. Isolated means the plates are joined to nothing, so Q, the charge, stays fixed. In panel a of the figure, the slab, marked K, of thickness t, sits in the gap d, and it makes the capacitance rise. Now U, the stored energy. U is one half C V squared, and C has gone up. So the stored energy goes up too, correct? Oh no, careful here, that is the trap, because V did not stay the same. V is Q over C, so with Q fixed, V falls as C rises. Right? So use the form with the quantity that stays fixed: U equals Q squared, over two times C, okay? Q is fixed and C goes up, so U goes down. With me? And the lost energy is not wasted. The field itself pulls the slab into the gap, and the energy the capacitor loses is the work it does on the slab. Clear? So the answer is no: the stored energy decreases. Always ask first what stays fixed. Got it?

7:40Quick revision: E and V, field lines, Gauss, C and energy

So, quick revision. Zero field need not mean zero potential, and an empty cavity sits at the potential of the conductor. The closed integral of E dot d l is zero, so there are no closed field lines, and Gauss's law holds for every closed surface. Capacitance depends only on geometry and the medium, and for an isolated capacitor, U equals Q squared, over two times C, which falls as C rises. Now try the practice question on this page. All the best!