Electric Charge and Coulomb's Law

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a metal ball and give it a positive charge. Is the ball now heavier, lighter, or exactly the same? Most students say the same, right? Charge is not mass. We will settle it very soon. And look, this is where all of electrostatics begins, and questions on Coulomb's law come up often in JEE papers. So, let's begin.

0:27Electric charge: what it is

So, what is electric charge? It is the property of matter that makes it exert electric forces, and also feel them. Okay? Charge is a scalar, so it has no direction. And it comes in two kinds, positive and negative. Like charges repel, and unlike charges attract, right? And the S I unit of charge is the coulomb, written as a capital C on screen. Clear?

0:52Quantised, conserved, additive

Now, three properties. First, quantisation. Every observable charge is q equals n times e, where n is an integer, and e is one point six times ten to the power minus nineteen coulomb. Okay? So, a quick check. Can a body carry a charge of two point four times ten to the power minus nineteen coulomb? Divide by e. Two point four over one point six gives one point five, which is not an integer. So no, such a charge cannot exist on a body. If you said no, very good! Second, conservation. The total charge of an isolated system never changes; charge is only transferred. With me? And third, additivity: charges add algebraically, with their signs. Also, charge does not depend on speed, while mass does. Got it?

1:46Charging changes the mass, and induction

Now, back to our puzzle. How does a body become positive? Does it gain something, or lose something? It loses electrons, yes, exactly! So its mass decreases, very slightly, and our charged ball is lighter by a tiny amount, so the puzzle is solved. With me? And a negative body gains electrons, so its mass goes up, again by a tiny amount. Clear? Now, charging by induction. Bring a charged rod near a conductor, and it separates the charges inside the conductor. Okay? Next, earth the far side. Then remove the earth first, and only then the rod. The conductor is left with a charge opposite to that of the rod. So remember the order: earth off first, rod away second. Right?

2:35Coulomb's law

Now, the main law. Two point charges, q one and q two, where the small one and two are just labels, sit a distance r apart. The force F equals one over four pi epsilon naught, times q one times q two, over r squared. Epsilon naught, the Greek letter with a small zero, is the permittivity of free space. And we call that whole constant in front k. Okay? The value of k is nine times ten to the power nine, newton metre squared per coulomb squared. And epsilon naught is eight point eight five times ten to the power minus twelve, coulomb squared per newton per metre squared. The little minus signs in the powers mean per. Right? Now the conditions, and they really matter. This law is for point charges, or for spherically symmetric charges when we are outside them, and in this form, for vacuum. With me? And the vector form. F with an arrow on top, subscript two one, is the force on q two due to q one; the arrow means vector. It equals k times q one q two, over the modulus of r two minus r one, cubed, times the vector r two minus r one. Here r one and r two are the position vectors, and the vertical bars mean the modulus. Got it?

3:51Coulomb's law in a medium

Now, what if the charges sit fully inside a medium? Then the force becomes F naught over capital K, okay? F naught, with the small zero, is the force in vacuum, and capital K, also written epsilon with a small r, is the dielectric constant. But sometimes only a slab sits between the charges. Then a slab of thickness t, and dielectric constant K, acts like air of thickness t times root K, right? So the air left is r minus t, and the slab counts as t times root K, okay? F equals k times q one q two, over the bracket, r minus t, plus t times root K, whole squared. Clear?

4:31Superposition, and a classic trap

Now, more than two charges. The net force on a charge is the vector sum of the forces due to all the others, so directions matter. Okay? And each force is calculated as if the other charges were absent. Simple, right? So let me ask you this. Two charges are pushing on each other, and I bring a third charge close by. Does the force between the first two change? No, it does not, and this is a classic trap! The force between two charges is not changed by a third charge. What changes is the net force on each charge, because the third charge adds its own push or pull. Got it?

5:09Example 1: force between 2 and -3 microcoulomb at 30 cm

Alright, example one. Find the force between two microcoulomb and minus three microcoulomb charges, placed thirty centimetres apart in air. The Greek letter mu means micro, ten to the power minus six. Pause the video and try it first. Done? Okay, let's check. Thirty centimetres is zero point three metre, so r squared is zero point zero nine metre squared. Put thirty straight into the formula, and oh no, everything goes wrong! Always convert to metres. Okay? Now use the magnitudes, with the same k for air. F equals nine times ten to the power nine, times two times ten to the power minus six, times three times ten to the power minus six, over zero point zero nine. The minus sign only tells us the direction. Right? On top, nine times two times three is fifty four, and the powers of ten give ten to the power minus three. Fifty four times ten to the power minus three, over zero point zero nine, gives zero point six newton. The charges are unlike, so the force is attractive. And that is the full answer, zero point six newton, attractive.

6:22Example 2: new force with a dielectric slab between charges

Now a slightly harder one. Two point charges, one metre apart in air, repel with a force F, okay? Then a slab of thickness zero point five metre, and dielectric constant four, is placed between them. Find the new force. Pause the video and try it first. Alright, the dielectric constant is four, so the new force is F over four, right? No, that is exactly the trap, so be careful! F over K is for charges fully inside the medium, but this slab is only half a metre thick. So use the effective separation, r minus t, plus t times root K, which is one minus zero point five, plus zero point five times root four. Okay? Root four is two, and zero point five times two is one. So the half metre slab acts like a full metre of air, and the effective separation is zero point five plus one, which is one point five metres. With me? The force goes as one over r squared. So the new force, F prime, with the little dash, is F over one point five squared, that is F over two point two five, which equals four F over nine. Clear?

7:38Quick revision: charge properties, Coulomb's law, slab rule

So, quick revision. Charge is quantised, q equals n times e, conserved and additive, and a positive body has lost electrons. Coulomb's law, F equals k times q one q two over r squared, is for point charges in vacuum; in a medium, it becomes F naught over K, okay? For a slab, use r minus t, plus t times root K, and for many charges, take the vector sum; the force between any two is unchanged. Now go and try the practice problems on this page. All the best!