1. Three point charges are arranged in a horizontal line as shown below. Find th

Question

1. Three point charges are arranged in a horizontal line as shown below. Find the electric forces (in units of kQ2/R2) on the charges given the following: Q1 = -48 Q, Q2 = -80 Q, Q3 = -64 Q, r1 = 4 R, and r2 = 4 R. Remember that a positive force points to the right and a negative force points to the left.

What is the net force on charge Q1?

What is the net force on charge Q2?

What is the net force on charge Q3?

What is the sum of the forces on all three charges?

Choose true or false for each statement below about the uniform electric field shown above.

true/false  The plate on the right is positively charged
true/false  The electrostatic force on a charge is greater at point A than at point B
true/false  A negative charge at Point A is at a higher potential than point B
true/false  The kinetic energy of a negative charge moved from point B to point A will increase.
true/false  A positive charge moved from point A to point C will have a negative change in potential energy.

3. The figure below shows two long parallel, straight wires carrying currents, I1 = 2 A and I2 = 10 A. The distance between the wires is 8 m. What is the magnitude of the magnetic field (in nT) at point A , which is at a distance of 12 m from wire #1?

4. A light beam goes from a medium with index of refraction 1.7 into another medium with index of refraction 2.9.
a. What is the critical angle of total internal reflection?
20.2°
35.9°
36°
None of the above.

b. What is the critical angle of total internal reflection if the beam direction is reversed, i.e., if the beam goes from the medium with index of refraction 2.9 into the medium with index of refraction 1.7?
20.2°
35.9°
36°
None of the above.

Explanation / Answer

part 4) When light is incident upon a medium of lesser index of refraction, the ray is bent away from the normal, so the exit angle is greater than the incident angle. Such reflection is commonly called "internal reflection". The exit angle will then approach 90° for some critical incident angle .

n1*sinthetac = n2*sin90

part 1 )

n1 = 1.7

thetac = sin^-1(2.9/1.7)

this is not possible because sin value varies from (-1 to 1)

d option

part b )

here n1 = 2.9

n2 = 1.7

theta_c = sin^-1(1.7/2.9)

theta_c = 35.9 degree

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