Electromagnetics, Waves, and Optics - Physics

In Coloumb's law, an increase in both interacting charges will cause an increase in the magnitude of the electrical force between them. Specifically if the magnitude of both interacting charges is doubled, this will quadruple the electrical force.

Because of the principles of superposition, each electric force that acts from the charges at the corners on to the charge at the center can be broken into components. Since all the charges are positive, all the forces will be repulsive. The forces acting from the top left and bottom right corners will cancel, leaving only the repulsive force coming from the bottom left corner.

Positive charges in an electric field will experience an electric force that is in the same direction as the electric field. If the charge is negative, the force will be in the opposite direction of the electric field. Since we are talking about an electron moving in an electric field that points in the negative y-direction, the electron will feel a force that points in the positive y-direction, or upwards.

When the negatively charged rod is brought near one of the two spheres, the presents of the negative charge will induce a flow of charge in the spheres such that regions farthest away from the charged rod will become most negative and regions near the rod will become most positive. This is called charge by induction.

Charge by induction happens when a charged object is brought in the vicinity of a neutral object. The presents of the charged object will cause the free charges in the neutral object to shift such that the neutral object becomes polarized. When the charged object is positive, this will induce a negative charge on a neutral object.

Positive charges in an electric field will experience an electric force that is in the same direction as the electric field. If the charge is negative, the force will be in the opposite direction of the electric field. Since the charged particle experiences a force which is opposite to the electric field, the sign of the charge must be negative.

A conductor is defined as a object free to move charges. In particular, valence electrons, which are the outer most electron in each atom and the most free to move, travel inside the conductor until the net electric field inside the conductor is zero. These electrons will move until this condition has been met. Because of the presents of charged particles at the surface and the condition that they are no longer moving, any electric field at the surface must be perpendicular to that surface.

Gauss' laws states that the the electric flux through a Gaussian surface will be proportional to the net total charge inside the Gaussian surface. By inspection of the figure, we see that Gaussian surface B is the correct answer.

According to Gaussian law, the electric flux will be zero when the net electric charge inside the Gaussian surface is zero. By inspection, we see this is Gaussian surface C.

According to Gauss's law, the total electric flux is equal to the net total electric charge inside the a Gaussian surface. If the Gaussian surface is three times larger, the electric flux will be the same if both Gaussian surfaces contain the same amount of total electric charge.

The direction of electric field describes the trajectory of a positive test charge. Since a positive test charge would want to move away from the positive metal plate and towards the negative metal plate, the direction would be A.

For this question, we're given the definition of near point. We're told what the near point is in the average person, and also the near point for a certain person who can't see well and needs glasses. We're asked to find the refractive power of a lens that will bring this individual's near point to the average, healthy value.

As was stated in the question stem, the near point is the closest distance of an object from the eye where that object can still be seen clearly. In the question, we're told that the normal value for this is . Moreover, a person with a near point of means that the object will need to be twice as far away, and no closer, to be seen clearly. Thus, in order to correct for this, a lens will be needed.

The idea is to be able to make the individual see things clearly when objects are located away. To accomplish this, a lens will need to diffract the light coming from an object away. This diffracted light will then need to form an image away, which is where this particular individual's near point is.

With this information in hand, we can use the lens-makers equation to solve for the refractive power of the lens.

We know that the object will be located a distance of away. Moreover, the image will need to form at a distance of away. However, since the image is forming on the same side of the lens that the object is located, the image will be virtual. Thus, the value used in the equation will be .

Also, remember that to find refractive power, we'll need to have our units be in meters.

Furthermore, recall that refractive power is equal to the inverse of focal length.

Hence, the power of the refractive lens will need to be diopters to correct this person's near point.

The important concept being tested in this question is the Doppler effect. When a source of sound waves (or any other wave) is emitted from a source and is moving relative to some observer, the actual frequency of the wave will be different for the observer.

For starters, we'll need to use the Doppler equation.

The trickiest part about this question is to decide whether to add or subtract in both the numerator and denominator. To find the right sign orientation, it's helpful to do a quick thought experiment. Imagine that only one of them is moving and the other is stationary. Decide how that will affect the observed frequency; will it increase it or decrease it? Then repeat for the other one.

First, let's consider the detector (aka observer). We're told that car B, the detector, is traveling ahead of car A and also in the same direction. This means that car B is driving away from car A. So in this situation, is the frequency that car B hears expected to increase or decrease? The answer is that it will decrease. Since car B is traveling away from car A, each successive wave will take longer to reach car B. Hence, we will use subtraction in the numerator because that will make the observed frequency smaller.

Now let's apply this same logic to the denominator, which deals with the source of the sound waves. We know that car A is traveling in the same direction as car B and is behind. This means that car A is traveling toward car B. So from this perspective, each successive wave is expected to get closer together, thus making the time between each wave smaller and the frequency bigger. In the denominator, will adding or subtracting make the observed frequency bigger? The answer is subtraction. By making a smaller number in the denominator, the entire fraction becomes larger.

Keeping this information in mind, we'll need to use subtraction in both the numerator and denominator. Once we plug in the values given in the question stem, we have everything we need to solve for the answer.

To begin, we'll start with the Doppler Effect equation:

Rearrange the equation by solving for to get:

Finally we plug in our known values and solve for :

For this question, we're told that light is reflected off of the surface of another medium. As a result, the light loses some of its energy, which is transferred into the medium where it reflected.

Because the light has lost some of its energy, we need to determine how the wavelength and frequency of the wave will be affected. To do this, we can recall the equation for the energy of a wave.

Where is Planck's constant and represents the frequency. We can see that when energy is decreased, the frequency also decreases. Thus, we can eliminate two of the answer choices.

To see how the wavelength changes, it's important to recall the relationship that wavelength and frequency have with each other.

Where is the speed of light and is the wavelength. Since the light is still traveling in the same medium, its speed will not change. Thus, as the frequency of the wave decreases, the wavelength has to increase. We can alternatively show how all of these variables are related as follows.

So, in summary, when the energy of light decreases, the frequency will decrease and the wavelength will increase.

Each harmonic has a wavelength that behaves according to the equation

, where is the wavelength and is the harmonic.

Since this is the 5th harmonic, and . Solving for wavelength gives .

For this question, we're told that a standing wave is contained within a pipe closed at both ends. We're also given the wavelength for the fifth harmonic, and are asked to find the length of the pipe.

The first step to solve this problem is to use the equation for a pipe closed at both ends.

Since we're told which harmonic the wave is on, as well as its wavelength, we have everything we need to solve for the length of the pipe.

Since the pipe is open at both ends, we can use the equation where f is the frequency we are solving for, is the number of the harmonic, is the velocity at which sound moves through air, and L is the length of the pipe.

We know from the question that:

When we plug in these values into the frequency equation, we get as the answer.

To solve this one will require two equations. First, to find focal length, you will need the lens equation.

But we are only given the object distance, not the image distance. So to find that we need to know the relationship between magnification and object and image distances.

Plug in the quantities we know.

(remember it was inverted)

. Then solve for which is . So the projector would sit that far from the board.

Now finally we can find the focal length of this lens.

Plugging in:

Solving for gives a focal length of or

For this question, we'll need to use the thin lens equation in order to determine the image distance from the lens.

Because this is a convex lens, the refracted light will form an image on the side of the lens where light is expected to go (real image). Thus, we know that the focal length should be a positive value. Moreover, the object distance is also positive.

Rearranging the above equation, we can isolate the term for image distance.

Then we can plug in the values given to us in the question stem.

And finally we take the inverse of this value to arrive at the correct answer.