L¨osningar till Typtenta Elektromagnetism och ljus Problem 1

Elektromagnetism och ljus

Problem 1

a) Reasoning: In the setup of this problem is entirely one-dimensional, you only have to think about what happens along the line of the two charges. To determine the electric field is pretty straight-forward once the formula is known.

The electric field is a vector quantity, the contributions from different charges add (vectorially). This means that you have to determine the electric field from each of the charges, and think about whether the two contributions add or subtract from each other. The electric field points away from positive charges, and towards negative charges. This is implied by the sign in the equation for the electric field, when you consider a negative charge the field will be negative which just means that it is pointing towards the source. You still have to think yourself whether this direction (towards the source) adds or subtracts with the electric field from other sources.

Solution: The electric field from a single point charge is E =

N/C, where k is Coulomb’s constant k =

= 9 · 10

. The electric field from q

is

E = kq

r

= 9 · 10

× 20 · 10

0.05

= 7.2 · 10

N/C

and since q

is positive the field point away from the source, i.e. to the right.

The electric field from q

is E = kq

r

= 9 · 10

× −5 · 10

0.05

= −1.8 · 10

N/C

where the negative sign means that the electric field points towards q

. Since q

is to the right of P , the field generated by the source q

in point P is pointing to the right. So the contributions from q

and q

in the point P point in the same direction, to the right. The field will therefore be stronger than from any of the two sources alone, and the magnitude of the two contributions have to be added. The total field in P is therefore 9 · 10

N/C and pointing to the right.

b) Reasoning: When the solution to a) is known, the force on a charge q which

is placed in point P is easy to calculate from the equation for the force acting on

a charge placed in an electric field. Regarding the sign of the test charge, you again have to use your reasoning (although it is not required for this particular problem). For a positive test charge, the force acting on it will point in the same direction as the electric field. For a negative test charge however, the force acting on it will be in the opposite direction of the electric field.

Solution: The force acting on a test charge q in an electric field E is F = qE N. So for our test charge q = −4 · 10

C we get

F = qE = −4 · 10

× 9 · 10

= −3.6 · 10

N

where the negative sign means that the force is pointing in the opposite di- rection from the electric field, i.e. to the left (it is attracted by the positive charge q

and repelled by the negative charge q

).

c) Reasoning: This one is maybe a bit trickier. First it is clear that the field is never zero anywhere in between the two charges q

and q

since they are of opposite charge and therefore the fields from them add together in between the charges as was seen in a). The the left of q

, there can also not be any point where the field is zero since the magnitude of q

is larger than q

and since q

will be closer to any point x which is to the left of q

there will always be a non-zero electric field to the left of q

. The solution must be some point x to the right of q

, where the electric field from q

point to the right and the electric field from q

point to the left (towards q

). We must find the point x where the longer distance from q

cancels the fact that q

has a larger charge than q

.

Solution: Let us say that the charge q

is located at x = 0, and we are looking for a point x which is to the right of q

. The distance r

from q

to the point x is then just x. The distance r

from q

to the point x is then 10 cm longer than r

, i.e. r

= x + 0.1 m. Now we require that the electric field contributions from q

and q

to the point x are of the same strength, since we have already reasoned that they contribute in opposite directions and therefore will cancel each other

:

E

+ E

= 0 ⇒ kq

(x + 0.1)

= −kq

x

⇒ q

x

= −q

(x + 0.1)

⇒ q

x

= −q

(x

+ 0.2x + 0.01) = −q

x − 0.2q

x − 0.01q

⇒ (q

+ q

)x

+ 0.2q

x + 0.01q

= 0

1

Here the equations are written such that the nagtive sign of the charge q

is respected. Per- sonally I often find it easier to first just reason about the directions of the contributions (whether they cancel or not) and then just work with the magnitude of the charges q

= |q

| and q

= |q

|.

The relation would just read E

= E

then, and you would end up with the same equation to solve

in the end.

So we are left with just a second-order polynomial, ax

+ bx + c = 0, to solve.

Solving it and the putting in the numbers for q

and q

gives:

x = −b ± √

b

− 4ac

2a = −0.2q

± p(0.2q

)

− 4(q

+ q

)0.01q

q

+ q

= 1 · 10

± p(−1 · 10

)

+ 3 · 10

30 · 10

= 1 ± √

4 30

We end up with two solutions (as always with a quadratic equation), but only one which has the positive x value that we are seeking

. So the final solution is that x = 3/30 = 0.1 m. So in a point 10 cm to the right of q

the total electric field is zero.

Problem 2

a) Reasoning: The potential generated by a single charge is easily calculated once the formula is known.

Solution: The potential generated by a point source is given by U =

V.

We know that k = 9 · 10

and that q = 2 µC. So U = 9 · 10

× 2 · 10

0.1 = 1.8 · 10

V.

b) Reasoning: The first part of the problem is trivial once the solution to a) is known. For the second part of the problem, we can make use of the hint. If the proton start at r

= 10 cm and moves to r

= 50 cm, it will have traveled from a point with higher potential to a point with lower potential. The potential U is the potential energy per unit charge, so by multiplying the potential with the charge of the proton we can find the potential energy in each of the two points. Since energy is conserved, the difference in potential energy has been transfered into kinetic energy (the initial kinetic energy in point r

is zero since the velocity is zero there). A similar situation is a falling object, it starts at rest in a point with high potential energy (i.e. high up) and after it has been falling some distance it has lower potential energy and the potential energy difference has been translated into kinetic energy. For the proton in our problem, once we have the magnitude of the kinetic energy (from the potential energy difference), we can calculate which velocity it corresponds to.

Solution: Let us call the potential we calculated in a) as U

, and then we

2

The solution with negative x corresponds to a point to the left of q

. This may be confusing,

but in that point E

and E

are also of the same magnitude, but the are pointing in the same

direction so the total field is not zero there. This is why it pays off reasoning about where the

solution should be first, and then maybe just work with the magnitude of the charges.

find the potential at r

= 50 cm:

U

= 9 · 10

× 2 · 10

0.5 = 3.6 · 10

V.

The potential energy for a charge q

in a potential U is E = Uq. The difference in potential energy for the proton with charge q

= 1.6 · 10

C is thus:

∆E = U

q

− U

q

= (18 · 10

− 3.6 · 10

) × 1.6 · 10

= 2.3 · 10

J The kinetic energy of the proton is equal to ∆E, and we can use the formula for kinetic energy to find the velocity v:

∆E = mv

2 ⇒ v = r 2∆E

m =

r 2 × 2.3 · 10

1.67 · 10

= 5.25 · 10

m/s

Problem 3

a) Reasoning: First of all, let’s say that text is correct and the figure is wrong when it comes to the currents in the two wires. The force can be found from the formula

=

, where µ

= 4π · 10

. The direction of the force can be found from the right-hand rule, the direction of the B-field at wire D goes down into the page which means that the force is towards wire C.

Solution: We are looking for the force on 1 m of wire, so ∆L = 1 m. The force is then:

F = µ

I

I

2πr = 2 · 10

× 4 × 8

0.1 = 6.4 · 10

N/m b) Reasoning: Finding this solution is easy once a) is done.

Solution: The force is the same as in a), but the direction is the opposite (the two wires repel each other).

Problem 4

a) Reasoning: Let us decide that the 200 V from the source is the r.m.s. of the voltage, which is what is most useful when doing calculation with alternating currents. Finding the current in the circuit is then easy from V=ZI.

Solution: The total resistance in the curcuit is R = 44 + 36 = 80 Ω. The total impedance can be found from the phasor diagrams (put X

upwards, R to the right and X

downwards and then calculate the vector sum which is Z). The length of Z can then be found from:

Z = p(X

− X

)

+ R

= √

60

+ 80

= 100 Ω

The (r.m.s. of the) current in the circuit is then I = 200/100 = 2 A.

b) Reasoning: The potential drop over each of the components can be found once we know the current from a) and using V=ZI again but keeping in mind that V

“points upwards”, V

“points to the right” and V

“points downwards”.

For the coil, which has both a V

and a V

component, you don’t have to cal- culate the resultant and the angle (even if it is possible of course).

Solution: Since the current is just 2 A: the voltage drop over the coil is V

= 180 V and V

= 72 V, the drop over the resistance is V

= 88 V and the voltage drop over the capacitor is V

= 60 V. Since V

and V

are completely out of phase (and therefore cancel each other), the total V is V = pV

+ (V

− V

)

= 200 V.

c) Reasoning: The effect factor is defined as cos φ, where φ can be found from φ = tan

.

Solution: We plug in the numbers:

φ = tan

 X

− X

R



= tan

 90 − 30 80



= 0.644 rad The effect factor is then cos φ = 0.8

d) Reasoning: The effect of the circuit is P = IV cos φ, so the effect is lower than pure IV (for a DC circuit) with the effect factor cos φ.

Solution: Lets plug in the numbers:

P

= IV cos φ = 0.8IV = 320 W

Problem 5

a) Reasoning: The distance of the image can be found from

+

=

. The focal point is defined by f = −

, where r is the radius of the mirror (so the focal point is -20 cm). The sign of the distances d

and d

are positive to the left of the mirror and negative to the right of the mirror. For convex mirrors, the distance of the image is always negative since the focal point is negative (so the image will be to the right of the mirror). The height of the image, h

, can be found from the magnification factor m =

= −

and the height of the object, h

.

Solution: We first calculate the distance d

for the image:

1 d

+ 1 d

= 1 f ⇒ 1

d

= 1 f − 1

d

= − 1 0.2 − 1

0.3 ⇒ d

= −12 cm The height of the image is found from the magnification factor:

h

= mh

= − d

d

h

= − −0.12

0.3 · 0.06 = 2.4 cm b) Reasoning: Read about how to do the graphical solution.

Solution: You have to draw it yourself on the paper.

Problem 6

a) Reasoning: The total effect is 3 · 10

W, or 3 · 10

J/s. Each photon carries some amount of energy, E =

where h is Planck’s constant h = 6.63 · 10

m

kg/s. We can then find how many photons we need.

Solution: The energy carried by one photon of wavelength 633 nm is:

E

= hc

λ = 6.63 · 10

× 3 · 10

633 · 10

= 3.14 · 10

J The total number of photons per second, n, is then:

E = nE

⇒ n = E E

= 3 · 10

3.14 · 10

= 9.55 · 10

b) Reasoning: The pressure is measured in force per unit area. Force is defined as F =

. The momentum change for each of the photons hitting the mirror is ∆p = 2p

(since they will now travel with momentum p

but in the opposite direction). The momentum of a photon can be found from p = E/c.

Solution: Let us first calculate the momentum for all the photons hitting the mirror during 1 s, which we can get from the total effect (remember the unit is J/s):

p = E/c = 3 · 10

3 · 10

= 10

⇒ ∆p = 2 · 10

kg · m s The force is now

F = ∆p

∆t = 2 · 10

N and the pressure is finally

P = F

A = 2 · 10

3 · 10

= 6.67 · 10

N/m