The magnitude of the force Two equal charges Q are placed a certain distance apart. They exert equal-and-opposite forces F on one another. Now one of the charges is doubled in magnitude to 2Q. What happens to the magnitude of the force each charge experiences? 1. Both charges experience forces of magnitude 2F. 2. The Q charge experiences a force of 2F; the 2Q charge experiences a force F. 3. The Q charge experiences a force of F; the 2Q charge experiences a force 2F. 4. None of the above.

The magnitude of the force Let’s examine this question from two perspectives. Newton’s Third Law – can one object experience a larger force than the other? 2. Coulomb’s Law – if we double one charge, what happens to the force?

The magnitude of the force Let’s examine this question from two perspectives. Newton’s Third Law – can one object experience a larger force than the other? No – the objects experience equal-and-opposite forces. 2. Coulomb’s Law – if we double one charge, what happens to the force?

Superposition If an object experiences multiple forces, we can use: The principle of superposition - the net force acting on an object is the vector sum of the individual forces acting on that object.

Worksheet – a one-dimensional situation Ball A, with a mass 4m, is placed on the x-axis at x = 0. Ball B, which has a mass m, is placed on the x-axis at x = +4a. Where would you place ball C, which also has a mass m, so that ball A feels no net force because of the other balls? Is this even possible?

Worksheet – a one-dimensional situation Ball A, with a mass 4m, is placed on the x-axis at x = 0. Ball B, which has a mass m, is placed on the x-axis at x = +4a. Where would you place ball C, which also has a mass m, so that ball A feels no net force because of the other balls? Is this even possible?

Worksheet – a one-dimensional situation Ball A, with a mass 4m, is placed on the x-axis at x = 0. Ball B, which has a mass m, is placed on the x-axis at x = +4a. Could you re-position ball C, which also has a mass m, so that ball B feels no net force because of the other balls?

Worksheet – a one-dimensional situation Ball A, with a mass 4m, is placed on the x-axis at x = 0. Ball B, which has a mass m, is placed on the x-axis at x = +4a. Could you re-position ball C, which also has a mass m, so that ball B feels no net force because of the other balls?

Worksheet – a one-dimensional situation Ball A, with a charge +4q, is placed on the x-axis at x = 0. Ball B, which has a charge –q, is placed on the x-axis at x = +4a. Where would you place ball C, which has a charge of magnitude q, and could be positive or negative, so that ball A feels no net force because of the other balls?

Worksheet – a one-dimensional situation Ball A, with a charge +4q, is placed on the x-axis at x = 0. Ball B, which has a charge –q, is placed on the x-axis at x = +4a. Where would you place ball C, which has a charge of magnitude q, and could be positive or negative, so that ball A feels no net force because of the other balls?

Worksheet – a one-dimensional situation Ball A, with a charge +4q, is placed on the x-axis at x = 0. Ball B, which has a charge –q, is placed on the x-axis at x = +4a. Where would you place ball C, which has a charge of magnitude q, and could be positive or negative, so that ball A feels no net force because of the other balls?

Worksheet – a one-dimensional situation Ball A, with a charge +4q, is placed on the x-axis at x = 0. Ball B, which has a charge –q, is placed on the x-axis at x = +4a. Could you re-position ball C, which has a charge of magnitude q, and could be positive or negative, so that ball B is the one feeling no net force?

Worksheet – a one-dimensional situation Ball A, with a charge +4q, is placed on the x-axis at x = 0. Ball B, which has a charge –q, is placed on the x-axis at x = +4a. Could you re-position ball C, which has a charge of magnitude q, and could be positive or negative, so that ball B is the one feeling no net force?

Worksheet – a one-dimensional situation Ball A, with a charge +4q, is placed on the x-axis at x = 0. Ball B, which has a charge –q, is placed on the x-axis at x = +4a. Could you re-position ball C, which has a charge of magnitude q, and could be positive or negative, so that ball B is the one feeling no net force?

A two-dimensional situation Simulation Case 1: There is an object with a charge of +Q at the center of a square. Can you place a charged object at each corner of the square so the net force acting on the charge in the center is directed toward the top right corner of the square? Each charge has a magnitude of Q, but you get to choose whether it is + or – .

Case 1 – let me count the ways. There is an object with a charge of +Q at the center of a square. Can you place a charged object at each corner of the square so the net force acting on the charge in the center is directed toward the top right corner of the square? Each charge has a magnitude of Q, but you get to choose whether it is + or – . How many possible configurations can you come up with that will produce the required force? 1. 1 2. 2 3. 3 4. 4 5. either 0 or more than 4

A two-dimensional situation Simulation Case 2: The net force on the positive center charge is straight down. What are the signs of the equal-magnitude charges occupying each corner? How many possible configurations can you come up with that will produce the desired force?

Case 2 – let me count the ways. There is an object with a charge of +Q at the center of a square. Can you place a charged object at each corner of the square so the net force acting on the charge in the center is directed straight down? Each charge has a magnitude of Q, but you get to choose whether it is + or – . How many possible configurations can you come up with that will produce the required force? 1. 1 2. 2 3. 3 4. 4 5. either 0 or more than 4

A two-dimensional situation Simulation Case 3: There is no net net force on the positive charge in the center. What are the signs of the equal-magnitude charges occupying each corner? How many possible configurations can you come up with that will produce no net force?

Case 3 – let me count the ways. There is an object with a charge of +Q at the center of a square. Can you place a charged object at each corner of the square so there is no net force acting on the charge in the center? Each charge has a magnitude of Q, but you get to choose whether it is + or – . How many possible configurations can you come up with that will produce no net force? 1. 1 2. 2 3. 3 4. 4 5. either 0 or more than 4

Worksheet: a 1-dimensional example Three charges are equally spaced along a line. The distance between neighboring charges is a. From left to right, the charges are: q1 = –Q q2 = +Q q3 = +Q What is the magnitude of the force experienced by q2, the charge in the center? Simulation

Worksheet: a 1-dimensional example Let's define positive to the right. The net force on q2 is the vector sum of the forces from q1 and q3. The force has a magnitude of and points to the left. Handling the signs correctly is critical. The negative signs come from the direction of each of the forces (both to the left), not from the signs of the charges. I generally drop the signs in the equation and get any signs off the diagram by drawing in the forces.

Ranking based on net force Rank the charges according to the magnitude of the net force they experience, from largest to smallest. 1. 1 = 2 > 3 2. 1 > 2 > 3 3. 2 > 1 = 3 4. 2 > 1 > 3 5. None of the above.

Ranking based on net force Will charges 1 and 3 experience forces of the same magnitude? Will charges 1 and 2 experience forces of the same magnitude (both have two forces acting in the same direction)?

Ranking based on net force Will charges 1 and 3 experience forces of the same magnitude? No, because both forces acting on charge 1 are in the same direction, while the two forces acting on charge 3 are in opposite directions. Thus, 1 > 3. Will charges 1 and 2 experience forces of the same magnitude (both have two forces acting in the same direction)?

Ranking based on net force Will charges 1 and 3 experience forces of the same magnitude? No, because both forces acting on charge 1 are in the same direction, while the two forces acting on charge 3 are in opposite directions. Thus, 1 > 3. Will charges 1 and 2 experience forces of the same magnitude (both have two forces acting in the same direction)? No, because one force acting on charge 1 is the same magnitude as one acing on charge 2, while the second force acting on charge 1 is smaller – it comes from a charge farther away. Thus, 2 > 1.

Ranking based on net force We can calculate the net force, too. If we add these forces up, what do we get? Is that a fluke?

Three charges in a line Ball 1 has an unknown charge and sign. Ball 2 is positive, with a charge of +Q. Ball 3 has an unknown non-zero charge and sign. Ball 3 is in equilibrium - it feels no net electrostatic force due to the other two balls. What is the sign of the charge on ball 1? 1. Positive 2. Negative 3. We can't tell unless we know the sign of the charge on ball 3.

Three charges in a line Ball 3 is in equilibrium because it experiences equal-and-opposite forces from the other two balls, so ball 1 must have a negative charge. Flipping the sign of the charge on ball 3 reverses both these forces, so they still cancel.

Three charges in a line What is the magnitude of the charge on ball 1? Can we even tell if we don’t know what Q3 is?

Three charges in a line What is the magnitude of the charge on ball 1? Can we even tell if we don’t know what Q3 is? Yes, we can! For the two forces to be equal-and-opposite, with ball 1 three times as far from ball 3 as ball 2 is, and the distance being squared in the force equation, the charge on ball 1 must have a magnitude of 9Q.

Let’s do the math. Define to the right as positive. Three charges in a line Let’s do the math. Define to the right as positive.

Two charges in a line The neat thing here is that we don't need to know anything about ball 3. We can put whatever charge we like at the location of ball 3 and it will feel no net force because of balls 1 and 2. Ball 3 isn't special - it's the location that's special. So, let's get rid of ball 3 from the picture and think about how the two charged balls influence the point where ball 3 was.

Two charges in a line Ball 2's effect on ball 3 is given by Coulomb's Law: Ball 2's effect on the point where ball 3 was is given by Electric Field :   The electric field from ball 1 and the electric field from ball 2 cancel out at the location where ball 3 was.

Electric field A field is something that has a magnitude and a direction at every point in space. An example is a gravitational field, symbolized by g. The electric field, E, plays a similar role for charged objects that g does for objects that have mass. g has a dual role, because it is also the acceleration due to gravity. If only gravity acts on an object: For a charged object acted on by an electric field only, the acceleration is given by: Simulation

Electric field lines Field line diagrams show the direction of the field, and give a qualitative view of the magnitude of the field at various points. The field is strongest where the lines are closer together. a – a uniform electric field directed down b – the field near a negative point charge c – field lines start on positive charges and end on negative charges. This is an electric dipole – two charges of opposite sign and equal magnitude separated by some distance.

Electric field vectors Field vectors give an alternate picture, and reinforce the idea that there is an electric field everywhere. The field is strongest where the vectors are darker. a – a uniform electric field directed down b – the field near a negative point charge c – field lines start on positive charges and end on negative charges. This is an electric dipole – two charges of opposite sign and equal magnitude separated by some distance.

Getting quantitative about field The field line and field vector diagrams are nice, but when we want to know about the electric field at a particular point those diagrams are not terribly useful. Instead, we use superposition. The net electric field at a particular point is the vector sum of the individual electric fields at that point. The individual fields sometimes come from individual charges. We assume these charges to be highly localized, so we call them point charges. Electric field from a point charge: The field points away from a + charge, and towards a – charge.

A triangle of point charges Three point charges, having charges of equal magnitude, are placed at the corners of an equilateral triangle. The charge at the top vertex is negative, while the other two are positive. In what direction is the net electric field at point A, halfway between the positive charges? We could ask the same question in terms of force. In what direction is the net electric force on a ______ charge located at point A?

A triangle of point charges Three point charges, having charges of equal magnitude, are placed at the corners of an equilateral triangle. The charge at the top vertex is negative, while the other two are positive. In what direction is the net electric field at point A, halfway between the positive charges? We could ask the same question in terms of force. In what direction is the net electric force on a positive charge located at point A?

Net electric field at point A In what direction is the net electric field at point A, halfway between the positive charges? 1. up 2. down 3. left 4. right 5. other

Net electric field at point A The fields from the two positive charges cancel one another at point A. The net field at A is due only to the negative charge, which points toward the negative charge (up).

Net electric field equals zero? Are there any locations, a finite distance from the charges, on the straight line passing through point A and the negative charge at which the net electric field due to the charges equals zero? If so, where is the field zero? 1. At some point above the negative charge 2. At some point between the negative charge and point A 3. At some point below point A 4. Both 1 and 3 5. Both 2 and 3 6. None of the above

Net electric field equals zero? Simulation Inside the triangle, the field from the negative charge is directed up. What about the fields from the two positive charges? Do they have components up or down? At the top, and at point A, the field is dominated by ____________. Far away, the field is dominated by ______________.

Net electric field equals zero? Simulation Inside the triangle, the field from the negative charge is directed up. What about the fields from the two positive charges? Do they have components up or down? Up. Thus, the net field everywhere inside the triangle has a component up. At the top, and at point A, the field is dominated by the negative charge. Far away, the field is dominated by the positive charges. In between, there must be a balance.

Worksheet: where is the field zero? Two charges, +3Q and –Q, are separated by 4 cm. Is there a point along the line passing through them (and a finite distance from the charges) where the net electric field is zero? If so, where? First, think qualitatively. Is there such a point to the left of the +3Q charge? Between the two charges? To the right of the –Q charge? Simulation

Where is the net field equal to zero? Is the net electric field equal to zero at some point in one of these three regions: to the left of both charges (Region I), in between both charges (Region II), and/or to the right of both charges (Region III)? The field is zero at a point in: 1. Region I 2. Region II 3. Region III 4. two of the above 5. all of the above

Worksheet: where is the field zero? In region I, the two fields point in opposite directions. In region II, both fields are directed to the right, so they cannot cancel. In region III, the two fields point in opposite directions. Now think about the magnitude of the fields.

Worksheet: where is the field zero? In region I, the two fields point in opposite directions. In region II, both fields are directed to the right, so they cannot cancel. In region III, the two fields point in opposite directions. Now think about the magnitude of the fields. In region I, every point is closer to the larger-magnitude charge than the smaller-magnitude charge, so the field from the +3Q charge will always be larger than that from the –Q charge. In region III, we can strike a balance between the factor of 3 in the charges and the distances.

Worksheet: where is the field zero? How can we calculate where the point is? If the point is a distance x from the +3Q charge, then it is (x – 4) cm away from the -Q charge. Define right as positive, so: The minus sign in front of the second term is not the one associated with the charge but the one associated with the direction of the field from the charge. The k's and Q's cancel. Re-arranging gives: Cross multiplying and expanding the brackets:

Worksheet: where is the field zero? Solve this using the quadratic formula to get two solutions: The two solutions are x = 2.54 cm and x = 9.46 cm. Which one is correct?

Two solutions Which of the two solutions is the one we want? 1. 2.54 cm 2. 9.46 cm 3. They are both valid solutions. Note: even if you decide one solution is not valid, you should be able to explain what its physical significance is.

Where is the field zero? The net electric field is zero 9.46 cm to the right of the +3Q charge (and 5.46 cm to the right of the –Q charge). The other solution is between the two charges, where the two fields point in the same direction. This point, 2.54 cm to the right of the +3Q charge, is where the two fields are equal.

A test charge A test charge has a small enough charge that it has a negligible impact on the local electric field. Placing a positive test charge at a point can tell us the direction of the electric field at that point, and tell us roughly how strong the field is. The force on a positive test charge is in the same direction as the electric field, because . Simulation

The net force on a test charge The diagram shows the net force experienced by a positive test charge located at the center of the diagram. The force comes from two nearby charged balls, one with a charge of +Q and one with an unknown charge. What is the sign and magnitude of the charge on the second ball? +Q/4 +Q/2 +Q +2Q +4Q none of these

The net force on a test charge This is the same as asking: If the net electric field at the point at the center of the diagram is in the direction shown, what is the sign and magnitude of the charge on the second ball? The vector is at a 45° angle, so the two forces (or fields) must be identical. The +Q charge sets up a force (or field) directed down. The second ball must set up a force (or field) directed left, away from itself, so it must be positive.

The net force on a test charge If the two forces (or fields) are the same, how does the magnitude of the charge on the second ball compare to Q?

The net force on a test charge If the two forces (or fields) are the same, how does the magnitude of the charge on the second ball compare to Q? It must be smaller than Q, because the second ball is closer to the point we’re interested in. The first ball is twice as far away. Because distance is squared in the equation, the factor of 2 becomes a factor of 4. To offset this factor of 4, the second ball has a charge of +Q/4.

The net force on a test charge, II The diagram shows the net force experienced by a positive test charge located at the center of the diagram. The force comes from two nearby charged balls, one with a charge of +Q and one with an unknown charge. What is the sign and magnitude of the charge on the second ball? +Q +Q ×√2 +2Q +2Q ×√2 +4Q none of these

The net force on a test charge, II In which direction is the force (or field) from the +Q charge? What are the possible directions for the force (or field) from the second ball?

The net force on a test charge, II In which direction is the force (or field) from the +Q charge? Down, away from the +Q charge. What are the possible directions for the force (or field) from the second ball? Left, if it is positive, or right, if it is negative. Can we combine a vector down with a vector left or right to get the vector shown?

The net force on a test charge, II In which direction is the force (or field) from the +Q charge? Down, away from the +Q charge. What are the possible directions for the force (or field) from the second ball? Left, if it is positive, or right, if it is negative. Can we combine a vector down with a vector left or right to get the vector shown? No – this situation is not possible.

Electric field near conductors, at equilibrium A conductor is in electrostatic equilibrium when there is no net flow of charge. Equilibrium is reached in a very short time after being exposed to an external field. At equilibrium, the charge and electric field follow these guidelines: the electric field is zero within the solid part of the conductor the electric field at the surface of the conductor is perpendicular to the surface any excess charge lies only at the surface of the conductor charge accumulates, and the field is strongest, on pointy parts of the conductor

Electric field near conductors, at equilibrium At equilibrium the field is zero inside a conductor and perpendicular to the surface of the conductor because the electrons in the conductor move around until this happens. Excess charge, if the conductor has a net charge, can only be found at the surface. If any was in the bulk, there would be a net field inside the conductor, making electrons move. Usually the excess charge is on the outer surface.

Electric field near conductors, at equilibrium Charge piles up (and the field is strongest) at pointy ends of a conductor to balance forces on the charges. On a sphere, a uniform charge distribution at the surface balances the forces, as in (a) below. For charges in a line, a uniform distribution (b) does not correspond to equilibrium. Start out with the charges equally spaced, and the forces the charges experience push them so that they accumulate at the ends (c).

A lightning rod A van de Graaff generator acts like a thundercloud. We will place a large metal sphere near the van de Graaff and see what kind of sparks (lightning) we get. We will then replace the large metal sphere by a pointy piece of metal. In which case do we get more impressive sparks (lightning bolts)? with the large sphere with the pointy object neither, the sparks are the same in the two cases

A lightning rod The big sparks we get with the sphere are dangerous, and in real life could set your house on fire. With the lightning rod, the charge (and field) builds up so quickly that it drains charge out of the cloud slowly and continuously, avoiding the dangerous sparks. The lightning rod was invented by __________.

A lightning rod The big sparks we get with the sphere are dangerous, and in real life could set your house on fire. With the lightning rod, the charge (and field) builds up so quickly that it drains charge out of the cloud slowly and continuously, avoiding the dangerous sparks. The lightning rod was invented by Ben Franklin.

Electric potential energy (uniform field) For an object with mass in a uniform gravitational field, the change in gravitational potential energy is: Similarly, for a charge q moving a distance d parallel to the electric field, the change in electric potential energy is:

Which way does it go? Whether it's an object with mass in a gravitational field, or a charged object in an electric field, when the object is released from rest it will accelerate in what direction? 1. Toward U = 0 2. Away from U = 0 3. In the direction of the field 4. In the direction of decreasing potential energy 5. In the direction of increasing potential energy

Which way does it go? Masses and positive charges behave in a similar way, but negative charges move opposite in direction to positive charges. In all cases, the object accelerates in the direction of decreasing potential energy. This is true whether the field is uniform or non-uniform. Simulation

Electric potential energy (for point charges) There is an electric potential energy associated with two charged objects, of charge q and Q, separated by a distance r. Note that the potential energy is defined to be zero when r = infinity. Potential energy is a scalar, so we handle signs differently than we do when we are handling vectors. Put the signs on the charges into the equation! This should remind you of the equivalent gravitational situation, in which: Electric potential energy :

Interacting point charges Case 1: a charge +q is placed at a point near a large fixed charge +Q. Case 2: the +q charge is replaced by a –q charge of the same mass. In which case is the potential energy larger? 1. Case 1 2. Case 2 3. neither, the potential energy is equal in both cases

Interacting point charges In case 1, the potential energy is positive. In case 2, the potential energy is negative. A positive scalar is bigger than a negative scalar (check with your bank manager about your bank balance if you have trouble with this concept!). Simulation Electric potential energy :

Interacting point charges We now release the charges from rest and observe them for a particular time interval. Assuming no collisions have taken place, at the end of that time interval which charge will have the greatest speed? 1. The +q charge 2. The –q charge 3. Both charges will have the same speed

Interacting point charges In this case, we can apply impulse – momentum ideas. The negative charge keeps getting closer to the central positive charge, so the force it feels increases. The opposite happens for the +q charge. Because the –q charge experiences a larger average force, its speed is larger after a given time interval.

Escape speed How fast would you have to throw an object so it never came back down? Ignore air resistance. Let's find the escape speed - the minimum speed required to escape from a planet's gravitational pull. How should we try to figure this out? Attack the problem from a force perspective? From an energy perspective?

Escape speed How fast would you have to throw an object so it never came back down? Ignore air resistance. Let's find the escape speed - the minimum speed required to escape from a planet's gravitational pull. How should we try to figure this out? Attack the problem from a force perspective? From an energy perspective? Forces are hard to work with here, because the size of the force changes as the object gets farther away. Energy is easier to work with in this case.

Escape speed Let’s do an equivalent problem for two charged objects. Find an expression for the minimum speed an electron, which starts some distance r from a proton, must have to escape from the proton. Assume the proton remains at rest the whole time. Let’s start with the conservation of energy equation. Which terms can we cross out immediately?

Escape speed Which terms can we cross out immediately? Assume no resistive forces, so Assume the electron barely makes it to infinity, so both Uf and Kf are zero. This leaves:

Escape speed If the total mechanical energy is negative, the object comes back. If it is positive, it never comes back. Solving for the escape speed gives: m is the mass of the electron; r is the initial distance between them. For an electron in the hydrogen ground state, we get vescape = 3.1 × 106 m/s.

Releasing two charges Simulation Two charged objects are placed close to one another and released from rest. Assume that each object is affected only by the other object.

Releasing two charges In the first case, we observe that the motion of one object is a mirror image of the motion of the other. What, if anything, can we say about the two objects? 1. They have the same mass. 2. They have the same charge (sign and magnitude). 3. Both of the above. 4. Neither of the above has to be true.

Releasing two charges How do the accelerations compare? How do the forces compare? (Can you answer this if you don’t know how the charges compare?) How do the masses compare?

Releasing two charges How do the accelerations compare? The accelerations are equal-and-opposite. How do the forces compare? (Can you answer this if you don’t know how the charges compare?) The forces are equal-and-opposite, even if the charges are different (Newton’s Third Law). How do the masses compare? They are the same, because m = F/a.