Electromagnetics, Waves, and Optics - College Physics

The formula for finding the voltage of a simple RC circuit is

, where is the capacitor voltage, is the source voltage, is the resistance, and is the capacitance.

We want to know when the capacitor will reach the voltage of the power source so

and thus

Using the properties of natural logs yields

Solving for yields

In this question, we're given a number of parameters of a circuit and are asked to find how we can use these various parameters to show the capacitance of the circuit's capacitor.

First, recall what a capacitor is; something that stores charge for a given voltage difference. In other words, when there is a voltage difference between the two plates of a capacitor, a certain amount of charge can be stored on these plates. The more charge that can be stored for a given voltage difference, the greater that capacitor's capacitance. This can be shown by the following equation.

From the above expression, is the capacitance which we are trying to find a proper expression for. is the charge accumulated on the plates, which is one parameter we're given. Voltage, , on the other hand, is not provided.

To put the voltage into different terms, we'll need to use Ohm's law, which states the following.

In other words, the voltage is proportional to both the current and the resistance of the circuit.

Since we are given both current and resistance as known parameters in the question stem, we can use these for our final answer. By substituting the in the capacitance expression with , we obtain the following answer.

Electric forces can be attractive or repulsive because charges may be positive or negative. In the case for gravitational forces, there are only attractive forces because mass is always positive.

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.

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 :

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.

The energy of a photon of a given frequency is determined by the equation

, where is Plank's constant

So plug in Plank's constant and the frequency of the x-ray photons to get an energy of very near

So this is all about the magnetic field strength around a current carrying wire.

The equation for this is:

But you must use Ohm's Law in order to find the current in the wire.

Since the wire has of resistance and the voltage through the wire is , that means the current in the wire is .

Being sure to change into , plug everything in and get the answer, which is

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.

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.