Ap Physics C Emag Example Questions Quick Guide

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AP Physics C: Electricity and Magnetism (EMag) is a challenging course that covers advanced topics in electromagnetism. This course is designed for students who have a strong foundation in physics and mathematics, and who are looking to expand their knowledge of electricity and magnetism. One of the best ways to prepare for the AP Physics C EMag exam is to practice with example questions. In this article, we will provide several example questions from the AP Physics C EMag exam, along with detailed explanations of how to solve them.

Example Question 1:

A charge of +2 microcoulombs is placed at the origin of an x-y coordinate system. A second charge of -3 microcoulombs is placed at the point (2,0) in the x-y plane. What is the magnitude and direction of the electric field at the point (3,3)?

To solve this problem, we need to calculate the electric field due to each charge separately, and then add the two fields vectorially. For the positive charge at the origin, the electric field is given by the equation:

E_positive = k*q/r^2

Where k is the Coulomb’s constant, q is the charge, and r is the distance from the charge to the point where we want to find the electric field. In this case, q = 2 microcoulombs and r = √(3^2 + 3^2) = √18.

Plugging in the values, we get:

E_positive = (9 x 10^9)*(2 x 10^-6)/(18) = 1 x 10^6 N/C

Next, we calculate the electric field due to the negative charge at (2,0). The distance from the negative charge to the point (3,3) is √(1^2 + 3^2) = √10. Using the same formula as before, we get:

E_negative = (9 x 10^9)*(3 x 10^-6)/(10) = 2.7 x 10^6 N/C

Finally, we add the two electric fields vectorially to get the total electric field at the point (3,3). Since the two fields are in opposite directions, we subtract the magnitudes and take the direction of the larger field:

E_total = 2.7 x 10^6 – 1 x 10^6 = 1.7 x 10^6 N/C (direction is towards the negative charge)

Example Question 2:

A square loop of wire with side length a is placed in a uniform magnetic field B pointing into the page. The loop is rotating about one of its sides with an angular velocity ω. What is the induced electric field in the loop?

To solve this problem, we need to use Faraday’s law of electromagnetic induction, which states that the induced electromotive force in a closed loop is equal to the rate of change of magnetic flux through the loop. In this case, the magnetic flux through the loop is changing because the loop is rotating.

The magnetic flux through the loop is given by the equation:

Φ = B*A*cos(θ)

Where B is the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop. Since the loop is rotating, θ is changing with time, and the induced electric field is given by:

E = -dΦ/dt = -dB*A/dt

Since the loop is rotating about one of its sides with angular velocity ω, the area of the loop is changing with time. The rate of change of the area is given by:

dA/dt = a*ω

Substituting this into the equation for the induced electric field, we get:

E = -dB*a*ω

Therefore, the induced electric field in the loop is proportional to the rate of change of the magnetic field and the angular velocity of the rotation.

Example Question 3:

Two parallel plates separated by a distance d are connected to a battery that maintains a constant potential difference V. What is the capacitance of the capacitor formed by the plates?

The capacitance of a capacitor is defined as the ratio of the charge stored on the plates to the potential difference between the plates. In this case, the potential difference between the plates is given by the battery voltage V, and the charge stored on the plates is equal to the capacitance times the potential difference:

Q = CV

Since the plates are parallel and separated by a distance d, the capacitance of the capacitor is given by the equation:

C = ε*A/d

Where ε is the permittivity of the material between the plates, A is the area of the plates, and d is the distance between the plates.

Therefore, the capacitance of the capacitor formed by the plates connected to the battery is ε*A/d.

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