Guass' Law and RC Solutions

In summary: I is the current and R is the resistance. Therefore, the charge on the capacitor will decrease over time according to the equation Q = Q initial exp(-t/RC), where Q initial is the initial charge on the capacitor.6. To determine the energy needed for a proton to orbit the Earth, you can use the equation E = qV, where q is the charge of the proton and V is the voltage. In this case, the energy will be equal to the kinetic energy of the proton, which is given by E = 1/2mv^2. Setting these two equations equal to each other and solving for V, we get V = 1/2mv^2/q. Plugging in the given values,
  • #1
skyliner34
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Guass' law and RC
1. Homework Statement

1. Find the potential due to a charged line of charge density lambda. the point of interest is on the axis of the line, and a distance afrom the near end of the lone . the line os of length b. the potential is: k lambda ln(a+b)/a).

2 Show that for two spheres connected by a conductor the ratio of the fields at the surface is given by E/e=r/R where one sphere has a radius R and Field E and the other sphere has a radius r and field e.

3. find the capacitances of two concentric spheres separated by a distance, d, small compared to their radius. show that the capacitance is approximatively the same as the capacitance of two parallel plates with the same area and separation: C= A(E)/d.

4. two capacitors 7.1x10^-6 farads and .000036 farads are connected in series and 10 volts is applied to the pair. the voltage is removed and the capacitors are connected in parallel, plus to plus. hat is the new voltage.

5. Show that the charge on a capacitor C connected in series with a resistor R is given by Q=Q initial exp(-t/RC)

6. you plan to build a large, cheap cyclotron using the Earth's magnetic fields(1 E-4 t) and orbiting just above the earth;s atmospheres(radius 5.9 E6 m) what energy in volts, should protons be given to just circle the earth? q= 1.6 E-19 m clb, m = 1.67 E-27 kg.

2. Homework Equations



3. The Attempt at a Solution
so far i only know i hv to apply gauss law and DC and RC stuffs
 
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  • #2


1. To find the potential due to a charged line, you can use Gauss' law and the equation V = kQ/r, where V is the potential, k is the Coulomb constant, Q is the charge, and r is the distance from the charge. In this case, you will need to integrate the equation over the length of the line to account for the varying charge density. The result should be k lambda ln(a+b)/a as stated in the problem.

2. For two spheres connected by a conductor, the ratio of the fields at the surface will be equal to the ratio of the charges on the spheres. This is because the conductor will redistribute the charges on the spheres such that they are in equilibrium. Therefore, the ratio of the fields will be given by E/e = Q1/Q2 = R/r, where Q1 and Q2 are the charges on the spheres, and R and r are the radii of the spheres.

3. To find the capacitance of two concentric spheres, you can use the equation C = Q/V, where Q is the charge on the spheres and V is the potential difference between them. In this case, the potential difference will be equal to the electric field multiplied by the distance between the spheres, which is d. Since the electric field is constant between two parallel plates, the capacitance will be approximately the same as that of two parallel plates with the same area and separation.

4. When two capacitors are connected in series, the total capacitance is given by 1/C = 1/C1 + 1/C2. In this case, the total capacitance will be 1/(7.1x10^-6 + .000036) = 7.1x10^-6 farads. When the capacitors are connected in parallel, the total capacitance will be the sum of the individual capacitances, which is 7.1x10^-6 + .000036 = .0000431 farads. Using the equation V = Q/C, the new voltage will be 10(.0000431/7.1x10^-6) = 61.2 volts.

5. The charge on a capacitor is given by Q = CV, where C is the capacitance and V is the voltage. In this case, the voltage will be changing over time due to the presence of the resistor. Using Ohm's law, V = IR,
 
  • #3


I am familiar with Gauss' Law, which is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is a powerful tool for solving problems involving the distribution of electric charges.

Regarding the first question, I would approach it by using Gauss' Law to determine the electric field at the point of interest due to the charged line. Then, I would use the definition of potential to calculate the potential at that point.

For the second question, I would use Gauss' Law again to determine the electric fields at the surface of the two spheres. Then, I would use the given information to calculate the ratio of the fields.

In the third question, I would use the formula for capacitance of parallel plates to approximate the capacitance of the two concentric spheres. This is because the distance between the spheres is small compared to their radius, making them behave similarly to parallel plates.

For the fourth question, I would use the formula for capacitors in series and parallel to calculate the new voltage after the capacitors are connected in parallel.

In the fifth question, I would use the relationship between charge, capacitance, and voltage in a RC circuit to derive the given equation for the charge on a capacitor.

Finally, for the sixth question, I would use the formula for the energy of a charged particle in a magnetic field to calculate the energy needed for a proton to orbit the Earth at a given radius and magnetic field strength.
 

1. What is Guass' Law and how does it relate to electricity?

Guass' Law is a fundamental law in electromagnetism that relates the electric field to the distribution of electric charges. It states that the electric flux through any closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space.

2. What is the significance of the RC circuit in electrical engineering?

The RC circuit is a basic circuit consisting of a resistor and a capacitor, often used in electrical engineering for filtering, timing, and signal processing. It is also used to model and analyze more complex circuits by breaking them down into simpler RC circuits.

3. How do you calculate the time constant in an RC circuit?

The time constant (represented by the symbol τ) in an RC circuit is calculated by multiplying the resistance (R) by the capacitance (C). It is a measure of how quickly the capacitor charges or discharges in the circuit.

4. How does the RC circuit behave in a DC circuit versus an AC circuit?

In a DC circuit, the capacitor in an RC circuit acts as an open circuit, preventing current from flowing through. In an AC circuit, the capacitor acts as a frequency-dependent resistor, allowing current to pass through and impacting the amplitude and phase of the signal.

5. What are some real-world applications of Guass' Law and RC solutions?

Guass' Law is used in various fields, such as electrical engineering, physics, and astronomy, to analyze and model the behavior of electric fields. RC circuits have many practical applications, including filtering noise in audio signals, regulating voltage in power supplies, and controlling the timing of electronic devices.

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