Physics 2220 Homework 5 Solutions

Physics 2220 Homework 5 Solutions ->->->-> __https://urluss.com/2tux6r__

Physics 2220 Homework 5 Solutions

This article provides the solutions to the homework 5 problems for the Physics 2220 course. The homework 5 covers topics such as electric potential, capacitance, and dielectrics. The solutions are based on the textbook University Physics with Modern Physics by Young and Freedman, 15th edition.

Problem 1

A point charge Q = +2.0 ÎC is located at the origin. A second point charge q = -5.0 ÎC is located on the x-axis at x = 0.3 m. Find the electric potential at the following points: (a) (0, 0.4 m), (b) (0.4 m, 0), and (c) (0.3 m, 0.4 m).

Solution

The electric potential due to a point charge q at a distance r from it is given by:

V = kq/r

where k = 8.99Ã10 NÂm/C is the Coulomb constant.

The electric potential at a point due to multiple point charges is the algebraic sum of the potentials due to each charge.

(a) The point (0, 0.4 m) is at a distance of 0.4 m from Q and at a distance of sqrt(0.3+0.4) = 0.5 m from q. Therefore, the electric potential at this point is:

V = kQ/0.4 + kq/0.5

V = (8.99Ã10)(2Ã10)/0.4 + (8.99Ã10)(-5Ã10)/0.5

V = 44.95Ã10 - 89.9Ã10

V = -44.95Ã10 V

(b) The point (0.4 m, 0) is at a distance of 0.4 m from q and at a distance of sqrt(0.4+0) = 0.4 m from Q. Therefore, the electric potential at this point is:

V = kQ/0.4 + kq/0.4

V = (8.99Ã10)(2Ã10)/0.4 + (8.99Ã10)(-5Ã10)/0.4

V = 44.95Ã10 - 224.75Ã10

V = -179.8Ã10 V

(c) The point (0.3 m, 0.4 m) is at a distance of 0.4 m from Q and at a distance of sqrt(0+0.4) = 0.4 m from q. Therefore, the electric potential at this point is:

V = kQ/0.4 + kq/0.4

V = (8.99Ã10)(2Ã10)/0.4 + (8.99Ã10)(-5Ã10)/0.4

V = 44.95Ã10 - 224.75Ã10

V = -179.8Ã10

Problem 2

A parallel-plate capacitor has a capacitance of 10 ÎF when there is air between the plates. The plates are connected to a 12 V battery. The separation between the plates is 2 mm. (a) How much charge is stored on each plate (b) How much energy is stored in the capacitor (c) A slab of dielectric material with a dielectric constant of 4 is inserted between the plates, completely filling the space. What are the new values of the capacitance, the charge on each plate, and the energy stored in the capacitor

Solution

(a) The charge on each plate is given by:

Q = CV

where C is the capacitance and V is the potential difference between the plates.

Q = (10Ã10)(12)

Q = 120Ã10 C

(b) The energy stored in the capacitor is given by:

U = 1/2 QV

U = 1/2 (120Ã10)(12)

U = 720Ã10 J

(c) The capacitance of a parallel-plate capacitor with a dielectric material between the plates is given by:

C' = kC

where k is the dielectric constant.

C' = 4(10Ã10)

C' = 40Ã10 F

The charge on each plate remains the same as before, since the battery maintains a constant potential difference. Therefore, Q = 120Ã10 C.

The energy stored in the capacitor with a dielectric material is given by:

U' = 1/2 QV'

where V' is the new potential difference between the plates.

V' = Q/C'

V' = (120Ã10)/(40Ã10)

V' = 3 V

U' = 1/2 (120Ã10)(3)

U' = 180Ã10 J ec8f644aee