AP Physics 1 & 2 Electric Fields & Forces

Hi everyone! We are thrilled to have you back with us here at Educator.com.0000

Today we are going to start a lesson on electric fields and forces which is the beginning of our unit on electricity and magnetism.0003

Our objectives are going to be to calculate the charge on an object and explain the Law of Conservation of Charge, describe differences between conductors and insulators, and explain the difference between conduction and induction.0010

We will also use Coulomb's Law to solve for the force on a charged particle due to other point charges, calculate the electric field due to one or more point charges and finally to analyze electric field diagrams.0023

Let us start by talking about electric charges.0034

As you know matter is made up of atoms, and those atoms even have smaller particles -- subatomic particles such as protons, electrons and neutrons.0037

Now protons have a charge of +1 and +1e is a +1 elementary charge, the smallest stable single charge, that is equal to 1.6 × 10^-19 coulombs (C), the SI unit of charge.0045

Electrons on the other hand have a charge of -1 elementary charge and neutrons are neutral.0060

Now most atoms are neutral -- they have equal numbers of protons and electrons.0066

The positives and the negatives balance out for a net charge of 0; it is neutral.0070

But if an atom loses its electrons, loses an electron or two, it is going to become positively charged, or if it gains an electron or two, it is going to become negatively charged.0075

We call those charged atoms 'ions'.0085

Now the fundamental unit of charge in the SI system is the coulomb (C), and that is a big amount of isolated charge.0089

The smallest isolated unit of electric charge is the elementary charge, 1e on a proton, -1e on an electron and it's magnitude is 1.6 × 10^-19 C.0098

Now like charges repel each other, opposites attract, and of course electric charge is conserved -- that is called the Law of Conservation of Charge.0110

You cannot spontaneously have just a (+) charge appear.0119

If you start off with 0 net charge, later on in a closed system you must still have 0 net charge.0122

If you start off with 0 net charge you could have a +1e and a -1e so that your net is still 0, but you cannot spontaneously get a +3e and a -1e for a net of 2 -- charge has to be conserved.0127

Now let us take a look at charge on an object.0143

Mittens the cat possesses an excess of 6,000,000 electrons.0146

Let us find the net charge on Mittens in coulombs SI units.0150

Well charge gets the letter (Q) and that is going to be 6 × 10⁶ electrons, and since they are negative I will put that there so we have -6 × 10⁶ elementary charges...0154

...and we are going to try and convert that to coulombs and the way we are going to do that is by multiplying by 1 again -- that old trick.0168

We want elementary charges to go away so I will put (e) down here and I want units of coulombs, so I will put that up here, and then I have to write numbers in here to make a ratio of 1.0175

Well 1 elementary charge = 1.6 × 10^-19 C.0186

Now when I go through and do the math, my elementary charge units are going to cancel out and I will get -6 × 10⁶ × 1.6 × 10^-19 C for a total of -9.6 × 10^-13 C as my answer. Great!0192

Let us take a look at the charge of an alpha particle.0216

An alpha particle, which is also known as a helium nucleus, consists of 2 protons and 2 neutrons, no electrons.0219

So what is the charge of an alpha particle?0225

Well if an alpha particle has 2 protons, those are the only charged particles, it must have a charge of +2 elementary charges.0228

But let us convert that into coulombs -- +2 elementary charges, and we need to multiply that by 1, I will write 1 as 1e = 1.6 × 10^-19 C...0239

...so I get a charge of 3.2 × 10^-19 C when I put that into SI units -- 3.2 × 10^-19.0254

All right. As we talk about materials -- conductors are materials that allow charges to move freely.0268

They have a very low resistivity, so charges can go through them very, very easily.0274

Insulators, on the other hand, do not allow charges to move freely.0279

They have what is known as a very high resistivity, and resistivity is a material property.0282

If you look up the resistivity of gold, it has a certain value; if you look up the resistivity of glass, it has a specific value.0288

It is a material property measured in ohm meters and it typically gets the symbol ρ, a squirrely little P.0296

All right. Materials may be charged by contact known as conduction.0306

You might have tried this trick before -- rub a balloon against your hair someday.0310

Some electrons from the atoms in your hair get transferred to the balloon.0313

The balloon now has a net negative charge and your hair, because it lost some electrons, now has a net positive charge.0317

You can also charge conductors by contact.0324

For example, if you bring a charge conductor into contact with an identical neutral conductor, you will share the charge across those two conductors.0327

Let us take a look and see how that would work.0335

A conductor carrying a net charge of 8 elementary charges is brought into contact with an identical conductor with no net charge.0338

When they are brought into contact because their conductors and charges can move freely, those charges -- those 8e -- they are all positive; that is a positive charge; they are going to repel; they want to get as far apart from each other as they can.0346

So what do they do? They split up so you get 4e on each conductor.0357

If those conductors are identical, then you get the exact same charge on each.0361

Then if you split them apart, you now have 4 elementary charges on each object.0365

So what would the charge be?0370

4e on each one, is going to be 4 times the charge on an elementary charge -- 1.6 × 10^-19 C, which is 6.4 × 10^-19 C.0372

Now on a conductor, the charge is always going to sit on the outside surface.0387

So the electric field inside the conductor, as we talk about fields here soon, is always going to be 0.0392

Something to remember is that an electric field inside a conductor is 0.0398

Let us talk about the electroscope.0405

The electroscope is a really cool tool that is used to detect small electric charges based on conduction.0407

It consists of a conducting rod in a beaker that is then insulated from the outside world, except for this metal knob that sticks out the top.0413

If you were to bring something like a positively charged rod over to it to share the charge, you are going to get a net positive charge on your metal rod and as that is distributed throughout the rod, at the bottom you have these two very thin leaves.0421

If you have positive charges on each of the leaves, they are going to repel and you get a spreading of the leaves, which indicates that you have a charge.0434

You could have the same basic thing happen if you were trying to do this with a negative charge.0442

If we draw just the metal part of our electroscope here and we bring a negatively charged rod near it...0446

...well now the electrons that are already on that metal rod without even touching, they are going to be repelled from the negative charge there and they are going to try and hang out as far away as possible from that negative charge -- leaving this end positive.0456

Well once again down here on those leaves you have negative charges on each leaves they will repel, so this is really a charge detector.0470

You could also charge by induction -- that is charging a conductor without actually coming into contact with another charged object.0480

So if we start off with a neutral electroscope here and bring a positive rod near it, we will have the electrons all want to gather near the positive charge -- opposites attract -- leaving a net positive charge down near the leaves of the electroscope; they will spread apart.0487

Now what we are going to do though is while we hold that rod in place, we are going to connect this metal bar to ground and by some connection to ground -- to the earth through a conductor -- the earth acts as an infinite sync or source of electrons.0502

If you need some electrons to balance things out you could pull them straight from the earth; we have tons of extras.0517

If you need to get rid of some, you can throw them into the earth very easily.0522

So now when you connect this to ground it sees all the positive charges here and you will start sucking some electrons up from the ground in order to make that balance nice and happy.0526

Now if you disconnect the ground connection, those charges are stuck on the metal conductor, you have a net negative charge on the electroscope and of course down here where you once again have negative charges on the leaves, the light charges repel each other and you see the spreading of the scope.0535

You have therefore charge to the electroscope by induction -- you have not touched it specifically to the source of charges to that glass rod.0553

Let us talk about electrostatic attraction.0564

A positively charged glass rod attracts object X. What can you say about the net charge of object X?0566

Well before we can answer that, let us think about what could happen here.0574

If we have a positively charged glass rod, it is pretty easy to see that if you have some other negatively charged object that it will attract.0577

So without a doubt, you could say that you could attract a negatively charged object.0590

That is easy, but what if the object is neutral?0594

Well if it is a conductor and it is a neutral object, you will actually have some of the electrons in the object who will want to hang out over near the positive charge.0598

It remains electrically neutral, but because you have a negative closer to the positive, then the positive-positive you have a net attractive force, so you can actually attract neutral objects.0607

What on the other hand would happen if you had that positive rod and instead of a conducting sphere like we had there let us talk about an insulating sphere.0618

Well the atoms actually in this insulating sphere are made up of a positive nucleus and a negative electron orbiting it or multiples.0629

When you bring the positive charge near it, the electrons are going to want to be attracted to that positive rod.0642

They will want to spend just a little bit more time in their orbits over toward that positive rod side.0647

So you are going to actually get the atoms, polar atoms or molecules to polarize just a little bit the ones at the edge, and as they do that you have a slight attractive force that is stronger between the positive and negative than the repulsive force from the positive to the positive.0652

You have created a slight polarization of those molecules electrically and you can attract a neutral insulator that way as well.0670

So what is the answer?0679

The net charge of object X could be neutral or it could be negative to get attracted.0680

Same thing would happen if you have a negative glass rod, it could attract something that was either positive or neutral.0686

The only way to prove that two objects are charged is by repulsion.0693

You can attract neutral objects with one charged object; you cannot repel unless both objects are charged.0698

Let us talk about Coulomb's Law as we try and quantify this electrical force of attraction or repulsion.0706

As we know like charged objects repel and opposites attract.0712

Charged objects therefore must apply a force upon each other -- that is known as the electrostatic force or the coulombic force.0717

And similar to gravity, the force of attraction or repulsion is determined by the amount of charge instead of mass and the distance between the charges.0723

So the electric force is equal to some constant (k) -- that is called the electrostatic constant, it has the value 9 × 10⁹N-m² per C² × the charge on the first object × the charge on the second object divided by the square of the distance between them.0732

It is another inverse square law -- and notice how similar that is to the force equation we had for gravity.0749

Universal gravitational force was (g) -- m1m2/r².0756

Notice what we have here. Instead of (k), we had (g) for gravity and instead of charge we have masses and the distance between their centers.0765

It is almost the same thing, just in a different regime.0777

The biggest difference is that gravity cannot repel, while the electrostatic force can attract or repel.0781

So determining the electrostatic force -- three protons are separated from a single electron by a distance of 1 micron or 1 × 10^-6 m.0790

Find the electrostatic force between them. Is that attractive or repulsive?0800

Well let us answer the easy question first.0805

If we have 3 protons, that is going to be a charge of +3 elementary charges, the electron will be -1 elementary charge, and opposite charges are going to attract.0807

Now, however, let us find the force between them.0819

Let us call our protons our first charge (q1), so that is going to be 3e or 3 × +1.6 × 10^-19 C, which is 4.8 × 10^-19 C.0823

Our second charge (q2) is going to be -1 × our elementary charge (1.6 × 10^-19 C) or -1.6 × 10^-19 C.0836

And the distance between them? R = 10^-6 m.0850

So now we can apply Coulomb's Law.0854

The electric force = k (the electrostatic constant) times the product of our charges divided by the square of the distance between them.0857

So that is going to be 9 × 10⁹N-m²/C² (k) × q1 (4.8 × 10^-19 C) × q2 (1.6 × 10^-19 C) all over the square of the distance between them -- 10^-6 m².0866

Now I want you to notice something here -- I did not worry about the signs when I am using Coulomb's law.0889

Although you can, it is usually a lot simpler to calculate the magnitude of the force and then use what you know about positive - negative charges, attraction or repulsion to figure out the direction of the force.0896

Typically, do not worry about the signs of the charges.0908

It is a lot easier to just figure it out using common sense once you have done your calculations.0911

This implies then that the electric force, when I put all of this into my calculator, comes out to be right around 6.9 × 10^-16 and the units of force of course are Newton's (N) and we already determined that that was an attractive force.0917

Let us take a look at a deflecting electron beam.0937

If we have a beam of electrons, a bunch of electrons all negative -- e-, e-, my symbol for electron (e-) -- all moving to the right.0940

It is directed into the electric field between two oppositely charged parallel plates.0951

What is the direction of the electrostatic force exerted on the electrons by the electric field?0956

Well even though we have not talked about electric field formally yet, it is pretty easy to see what is going to happen to our electron beam.0960

If these are negative charges, they are going to get attracted by this side so we are going to have a force that way to do attraction here, and they are going to be repelled by the negatives here so we will have a force that way.0967

So our net force -- both of these are going up, it is going to deflect it up, so we have a net force upwards.0978

Electric fields -- The electrostatic force, just like the gravitational force is a non-contact or field force -- we cannot see it.0989

The way we can detect it is by placing a charge somewhere in space and then seeing what force it has acting upon it.0997

The property of space that allows a charged object to feel a force is what we call the electric field -- that mental construct to help us understand what is going on.1005

You detect the presence of an electric field by placing a positive test charge at various points in space and measuring the force the test charge feels.1012

What that means is the electric field is always going to point in the direction of the force on a positive charge, not on a negative charge.1021

The electric field points the direction of a force on a positive charge.1029

We quantify this with the electric field strength vector (E) and that is the amount of electrostatic force observed by a charge per unit of charge.1034

So the electric field strength (E) is the electric force divided by the amount of charge; it is the force per unit charge.1044

Its units are going to be -- Well, force is Newton's (N), (q) charge is in coulombs so that is Newton's per coulomb (N/C) and we will later find out that that is equivalent to volts per meter.1052

Newton's per coulomb or volts per meter is the same thing.1066

And again, the direction of the electric field vector is in the direction a positive charge would feel a force.1070

So let us take a look at another example -- finding the field between some metal plates.1078

Two oppositely charged parallel metal plates at 1 cm apart exert a force of 3.6 × 10^-15N on an electron placed between the plates.1083

Calculate the magnitude of the electric field strength between the plates.1094

Well let us write down what we know so far.1098

First off, our parallel plates are 1 cm apart -- so the distance between our plates is 0.01 m and there is a force (an electric force) of 3.6 × 10^-15N on an electron.1100

What do we know about an electron?1115

Its charge is -1e or -1.6 × 10^-19C.1117

We want the magnitude of the electric field strength, so we are looking for (E).1124

Well from our definition, the electric field strength is the electric force divided by the charge, which is 3.6 × 10^-15N/-1.6 × 10^-19 C...1130

...which gives me an electric field strength of 2.25 × 10⁴ N/C and that would be negative, but since it is asking us for the magnitude, our answer is just 2.25 × 10⁴ N/C.1145

Let us talk a little bit more about the electric field and how we might be able to visualize it.1167

Since you cannot actually see it, you can visualize it by drawing what we call field lines that show the direction of the electric force on a positive test charge.1172

The electric field lines always point away from positive charges and toward negative charges.1180

Electric field lines never cross each other; they intersect conductors at right angles to the surface of the conductor; and stronger fields have closer or denser lines; and finally field's strength and line density decreases as you move away from the charges.1186

So having giving you some of these basic rules for our electric field lines, let us take a look and see what they actually look like.1203

Here I have four different examples.1209

In the upper left I am showing a positive charge and the electric field lines radiating outward from that charge and because it is a positive charge, electric field lines go away from it.1212

You have a radial pattern and here you have a more dense lines in closer to the charge, so you have a stronger electric field.1222

The further away you get, you have less dense lines, you have a less strong electric field -- which we know of course because the electric field or electric force follows an inverse square law relationship.1231

Around a negative charge, you also have this radial pattern but now the field lines point in.1245

Lines point away from positive charges and end at negative charges.1251

Now down here I have a couple of dipoles -- two charges near each other.1256

If we find the electric field, what we are going to do is vector addition of the individual fields.1260

But if you have a positive and a negative beside each other -- well electric field lines go from positive to negative charges.1266

So we are going to get a pattern that looks kind of like this, and you could fill in more lines if you wanted to as you interpolate between them, and so on.1273

The key is that they go from positive to negative -- they never cross -- and they show the direction of the force on a positive test charge.1284

If I were to go take and put a positive charge right there in space, it is pretty easy to see that the net force it is going to feel is probably going to be somewhere in that direction.1291

If I were to go take a positive charge and put it right there, that yellow positive charge is going to feel a force to the right -- toward that negative charge.1302

The electric field line is telling you the direction of the force on a positive test charge.1309

Now over here on the bottom right, we have two positive charges making a dipole.1316

While electric field lines go away from positive charges and they do not cross, you end up with a pattern that looks kind of like this.1321

What is interesting here is if you were to take a positive test charge and put it exactly in the middle of those two, the forces on it are going to balance out so you get a net force of 0.1327

Right in the middle, you actually have a dead spot while you are exactly at that point.1340

So that is a little help to get you going on visualizing electric field lines.1345

Let us talk about calculating the electric field due to a point charge.1352

Electric fields are caused by charges.1356

The electric field due to a point charge can be derived from what we know about the electric field and Coulomb's Law.1358

We already know that the electric field strength is the force divided by the charge and Coulomb's Law says that the electric force on two charged objects is this constant (k) times the product of the charges divided by the square of the distance between them.1363

We can put those together to find a definition, or a formula for the electric field due to a point charge.1379

If the electric field strength is force divided by charge and force is (k)q1q2/r²...1385

...and I still have this (q) in the denominator there, well we can cancel out or make a ratio of 1 from one of our sets of charges and we find that the electric field strength is kq/r².1397

That is the formula for the electric field due to a point charge.1412

To find the electric field due to multiple point charges, take the vector sum of the electric fields due to each of the individual point charges-- you just keep adding them up in a vector fashion.1425

So let us take a look at comparing electricity to gravity again, there are so many similarities.1436

Force is F = (k)q1q2/r² for electricity and for gravity it is (g)m1m2/r².1442

We have swapped the (k) for (g) and the (q) for (m), charge for mass, but the same basic pattern though.1449

The field strength -- electric field strength is the electric force divided by the charge, the gravitational field strength (g) was the gravitational force divided by the mass -- again a nice parallel.1456

The field strength due to a point charge was kq/r² and for gravity that was gm/r² -- the same parallel.1467

Now they have different constants and notice (k) here on the left is 8.99 × 10⁹?1477

I tend to just make that a nice, simple 9 × 10⁹, so if you see that written differently somewhere, do not worry, it is close enough.1482

On the right side the gravitational constant is 6.67 × 10^-11N-m²/kg².1490

Again, these are really just fudge factors to make the units work out and the charge units were coulombs; the mass units were kilograms -- a lot of parallels.1497

Let us take a look at a little bit more in-depth problem solving.1508

Find the electric field at the origin due to the 3 charges shown in the diagram.1513

We have a +2 C charge here at (0,8), a +1 C charge here at (2,2) and a -2 C charge at (8,0).1517

Well what we are going to do -- our strategy is going to be to find the electric field at the origin to each of the individual charges and then add those up in vector fashion.1527

So let us start by looking at our green charge up here at (0,8).1536

If that is a +2 charge the electric field is going to point away from it, so we are going to have the electric field going in all these directions and at the origin, that must be pointing down.1541

So right away we know that we only have to worry about the (y) component.1551

Now the electric field strength is kq/r² where (k) is 9 × 10⁹ and our (q) is 2 C and the distance from the origin to our charge is 8 m.1556

So that is going to be 8² or 2.81 × 10⁸ N/C down.1573

And if I want to write that in bracket notation, my (x) component is 0 and my (y) component is going to be -2.81 × 10⁸ N/C because we know that must be pointing down.1582

So there is the electric field due to our green charge.1599

Let us do the red charge next.1603

The electric field due to that red charge is again kq/r² where (k) is 9 × 10⁹, (q) is -2C, but again we are not going to worry about signs at the moment -- let us get the magnitude and we will use common sense to figure out its direction...1605

...and its distance is 8 units from the origin, so 8² is going to give us the exact the same thing -- 2.81 × 10⁸ N/C is the magnitude.1624

But if this is a negative charge, electric field lines point in to negative charges.1634

So from the origin, we must be going toward the right due to that -2 C charge, so this would be 2.81 × 10⁸ N/C toward the right or in the positive (x) direction.1641

In bracket notation, I am going to write that as 2.81 × 10⁸ N/C and it has no (y) component.1653

So we have the electric field due to the green charge, the electric field due to our red charge, so now let us see what it is due to our blue charge.1663

If this is +1 C, it should be pretty obvious to see that the electric field at the origin will be pointing away from it -- that direction.1672

So we will have components in the (-x) and the (-y) direction here.1679

Let us find its magnitude -- E = kq/r² which is 9 × 10⁹, our charge is 1 C and now we have to get the distance from the origin to that charge.1684

Well to do that, I am going to make a right triangle here and if that is 2 units and that is 2 units, then by the Pythagorean Theorem, this must be the square root of 2² + 2².1699

So that is going to be over square root of 2² + 2² squared.1711

Or 9 × 10⁹/8, which is about 1.13 × 10⁹ N/C, but that is going to be down and to the left at an angle of 45 degrees because it has equal (x) and (y) components, which we can see just from the symmetry of the situation.1719

So that means that if I want to break this into (x) and (y) components, this is going to be equal to...1740

...Well we are going to have to have its magnitude 1.13 × 10⁹ N/C and the (x) component, we will multiply by the cos(45 degrees) and I know that is going to be negative.1745

And for the (y), it is also going to be -1.13 × 10⁹ × the sin(45 degrees), the (y) component.1760

And those are going to be the same -- sin(45) and cos(45) are the same thing.1770

When I run that through my calculator, I find that that is going to be about -7.95 × 10⁸ for the (x) and -7.95 × 10⁸ for the (y) -- all in Newton's per coulomb.1774

Now I want the total charge or the total electric field at the origin due to those three.1793

So for the total, all I am going to do is I am going to add up the (x) components and add up the (y) components.1800

My (x) components, I have 0 from the green charge, I have +2.81 × 10⁸ from the red charge and I have -7.95 × 10⁸ from my blue charge -- so that is the (x) component.1808

For the (y) component, I have -2.81 × 10⁸ (green) + 0 (red) - 7.95 × 10⁸ (blue), and again all of that is N/C.1830

So I will pull out my calculator and add up all my (x)'s and add up all my (y)'s and I end up with a total electric field of about -5.14 × 10⁸ for the (x) and -1.08 × 10⁹ for the (y) and all of that is in N/C.1850

A little bit more math here to get through it but really the same basic concept -- find the electric field due to each of those points and then sum them up to get the total electric field over here at the origin.1874

We have a component to the left, we have a component that is down so we are going to have something where we have a net electric field down in that sort of southwestern quadrant (Quadrant 3 of the graph).1888

Let us take a look at another problem.1903

Determine the (x) coordinate here on this line where the electric field is 0 using the diagram below.1906

We have a +2 C charge here on the right and then +1 C charge here on the left.1912

We are going to follow the same strategy again.1918

Let us take a look at the electric field due to that blue charge and we will call that E1 and that will be kq1/r².1921

When we do this, let us try and figure out roughly where our answer is going to be.1933

As I look down here, if I have a +1 C charge to the right, it is going to give me an electric field to the right; +2 C over here to the left is going to give me an electric field to the left.1938

And because +2 is stronger, I would expect that my 0-point is probably going to be somewhere a little closer to the +1 charge than it is to the +2 charge.1948

So I am going to make a guess and say that we are probably going to be somewhere over in that sort of region.1958

If we call this distance between our +1 C charge there at -6 and our point (r), then that means that over here, this distance must be 11 - r.1965

If I want to draw the electric field due to E2, due to this red charge, we will call that E2 is kq2 over -- and our distance now will be 11 - r².1987

We want to find out the point where those two -- where the net electric field is 0, so where those are going to be exactly equal in magnitude.2001

Our total electric field is going to be -- well we have our blue electric field over here -- (k) × q1 (1 C) divided by r² and we will have to subtract because this is going to the left -- the electric field due to the red charge.2011

That will be -(k) and the charge is 2, divided by 11 - r², and we want the point where that all is equal to 0.2036

First thing I can do here as I look at my problem is I see a simplification right away -- we can divide (k) out of all of that.2049

I could rewrite this then as 1/r² and I will move the red part to the right-hand side, which is going to be equal to 2/(11 - r)².2059

Let us extend that out, that will be 121 - 11r - 11r = -22r + r².2074

Now with a little bit of math here, if I multiply both sides by r², on the left-hand side I am going to get 1 = 2r²/121 - 22r + r²...2084

...or with some cross-multiplication -- 2r² = 121 - 22r + r².2100

It is starting to look at a quadratic equation, so let us get it in that form.2110

This implies then that 2r² - r² = r² + 22r - 121 = 0.2115

Now we can apply our quadratic formula.2126

Some of you have calculators that may do that automatically or you can go through the work there and what I find is that I get a value for r of 4.56.2130

It means if this is 4.56 then this distance, 11 - r, must be 11 - 4.56 = 6.44.2143

So determine the x-coordinate where the electric field is 0 using the diagram below.2154

If we want the x-coordinate, well now I just need to go back and finish this problem up.2160

To get the x-coordinate, that is going to be -- well if we start here at -6 and we add this distance r to it (4.56), I get an x-coordinate of about -1.44.2165

So it looks to me like our actual point for the correct answer should be at about -1.44 right about there and we will make our guesstimate go away.2179

We have found the x-coordinate where the electric field is 0.2191

Let us take a look at an example where we look at gravity and electricity.2199

A distance of 1 m separates the centers of two small charged spheres.2203

The spheres exert gravitational force (Fg) and electrostatic force (FE) on each other.2207

If the distance between the sphere's center is increased to 3 m -- so we are tripling the distance -- the gravitational force and electrostatic force, respectively, may be represented as...2213

This is an inverse square law problem again -- we have tripled the distance and because distance is squared, we are going to have a change by a factor of 9.2223

Now is that going to be 9 times greater force or 1/9 the force?2233

Well we are going to have a smaller force because they are getting further away.2237

Because they are both inverse square laws, our correct answer must be A -- 1/9 the gravitational force and 1/9 the electrical force.2241

We will have a last problem here.2254

In the diagram below, (P) is a point near a sphere that has a charge of -2 C.2256

What is the direction of the electric field at point (P)?2262

The way I would start this sort of problem is if we have a negative charge, let us draw some electric field lines.2266

Electric field lines point in to negative charges.2272

This line is just going to keep going so it looks like over here at point (P), we would have an electric field pointing that direction.2277

So the direction of the electric field at point (P) would be to the left as shown right there.2287

Hopefully that gets you a good start on electric fields and forces.2297

Thank you so much for your time. Look forward to talking to you soon.2300

Make it a great day everyone!2303