[PHYS 102] - Final Exam Guide - Comprehensive Notes for the exam (55 pages long!)

76 views55 pages

rubymarten89

29 Nov 2016

School

University of Illinois

Department

Physics

Course

PHYS 102

Professor

Kuehn David

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

Is a defined location in space around a charge or set of charges. E = f / q = kq / r^2. Magnitude: e = f / q (absolute value of q) Direction: direction is the same as for the force that a positive charge would feel at that location. Electric field direction from + and charges: away from + charge, towards charge, magnitude is equal! Superposition principle: total e field due to several charges = sum of individual e fields. This is the same approach as for force. Plane of charge: a large plane of charges creates a uniform electric field (constant magnitude & direction, small amount of charge = small electric field, large amount of charge = large electric field. Two charged planes: two large parallel planes with equal and opposite charge +q and q separated by a small distance. E field is uniform between plates, 0 everywhere else (i. e. , above and below the plates)

Related Questions

Goal Use the superposition principle to calculate the electricfield due to two point charges.

Problem Charge q1 = 7.00 ÂµC is at the origin, and charge q2 = -5.35ÂµC is on the x-axis, 0.300 m from the origin (Fig. 15.12).

(a) Find the magnitude and direction of the electric field at pointP, which has coordinates (0, 0.400) m.

(b) Find the force on a charge of 2.00 10-8 C placed at P.

(a) Calculate the electric field at P.
Find the magnitude of 1 with Equation 15.6.
E1 = 3.93 105 N/C
Vector 1 is vertical, making an angle of 90Â° with respect to thepositive x-axis. Use this fact to find its components. E1x = E1cos(90Â°) = 0
E1y = E1 sin(90Â°) = 3.93 105 N/C
Next, find the magnitude of 2, again with Equation 15.6.
E2 = 1.92 105 N/C
Obtain the x-component of 2, using the triangle in Figure 15.12 tofind cos ?.
E2x = E2 cos ? = (1.92 105 N/C)(0.600)
E2x = 1.15 105 N/C
Obtain the y-component in the same way, but a minus sign has to beprovided for sin ? because this component is directeddownward.
E2y = E2 sin ? = (1.92 105 N/C)(0.800)
E2y = -1.54 105 N/C
Sum the x-components to get the x-component of the resultantvector. Ex = E1x2x = 0 + 1.15 105 N/C
Ex = 1 N/C
+ E
Sum the y-components to get the y-component of the resultantvector. Ey = E1y2y = 3.93 105 N/C - 1.54 105 N/C
Ey = 2 N/C
+ E
Use the Pythagorean theorem to find the magnitude of the resultantvector.
E = 3 N/C

The inverse tangent function yields the direction of the resultantvector.
? = 4Â°

(b) Find the force on a charge of 2.00 10-8 C placed at P.
Calculate the magnitude of the force (the direction is the same asthat of because the charge is positive). F = Eq = E(2.00 10-8C)
F = 5 N

(a) Place a charge of -3.60 ÂµC at point P and find the magnitudeand direction of the electric field at the location of q2.

E = N/C

? = Â° counterclockwise from the positive x-axis

(b) Find the magnitude and direction of the force on q2.

F = N

? = Â° counterclockwise from the positive x-axis

I Don't know how to do this Please help Me. Thank you. The top partis how i figured out the rest i think its the missing informationto solve the bottom part.

[PHYS 102] - Final Exam Guide - Comprehensive Notes for the exam (55 pages long!)

Document Summary

Get access

Related textbook solutions

Physics: Principles with Applications

Related Documents

PHYS 102 Study Guide - Final Guide: Electric Dipole Moment, Electric Potential Energy, Magnetic Dipole

PHYS 102 Lecture Notes - Lecture 3: Electric Field

PHYS 102 Study Guide - Midterm Guide: Relative Permittivity, Electric Potential Energy, Dielectric

Related Questions