PHYS 208 Midterm: PHYS 208 TAMU PHYS208 Exam1Spring 16 Solutions

20 views9 pages

turquoisegnu128

31 Jan 2019

School

Department

Course

Professor

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

Document Summary

Total score (out of 100) v lys \0)1 a. B: (20 marks) a point charge q, = -80. 0 nc is located at (x,y) = (2. 00 m, 1. 00 m), and a point charge q2 = -60. 0 nc is located at (x,y) = (0. 00, -2. 00 m). Determine the net electric field, magnitude and direction , at the origin of the (x,y) coordinate system. E;: q = -2>0 t"\ (_ :::: ~cq~10 4 (gol/"\c) (_ g 1v1 )"l- z -{s. \ ;;l 9 n/ c_, (_9x10"\x"~ 0 ~c-) \ {__: ). l c gf ) ( b 0 n c) (_ ;;i. h-"l) 2. ~ r\ l ~ 2 q t-1/ c. ) "2. Determine the net electric field, magnitude and direction, at the origin of the (x,y) coordinate system. i a = +"o. ,c lo. = + ( q x 10"1_) ( czsonc) (_ .

Related Questions

1) Two point charges are on the x-axis. One charge, q1 = 10 nC, islocated at the origin, and the othercharge, q2 = 18 nC, is locatedat x = 9.0 m. What is the force on q2?

a) 20 nN in the positive x-direction
b) 20 pN in the positive x-direction
c) 20 nN in the negative x-direction
d) 20 pN in the negative x-direction
e) 180 nN in the positive x-direction

2) A proton is moving in an electric field. the direction of theacceleration of the proton is
a) In the direction the proton is moving
b) opposite the direction the proton is moving
c) perpendicular to the direction the proton is moving
d) in the direction of the electric field
e) opposite the direction of the electric field

3) A thin spherical shell of radius 20 cm has 5.0 Î¼C of chargeuniformly distributed over its surface. What is the electric field10 cm from the center of the sphere

a) 4.5 x 10^6 N/C inward

b) 4.5 x 10^6 N/C outward

c) 2.3 x 10^6 N/C inward

d)2.3 x 10^6 N/C outward

e)0

Hi I need help to solve this problems it would really help me tostudy, hopefully someone can help me with as much details aspossible. (formuas, explanations, etc) thank you in advance!

1. A 50.0 kÎ© resistor is connected with a 30.0 Î¼F capacitor and a15.0 V battery with negligible internal resistance. Find the timefor the capacitor to be charged from 0 to 75% of its maximumcharge.

2. A point charge q₁ =5.00nC is placed at x=4.00cm,and a second point charge q_{2=-3.00 nC is placed at x= 6.00cm.}

a)What is the magnitude and direction of the electric field dueto the charges, at the origin x= 0cm?

b) What is the magnitude of the electrical potential at theorigin?

c) What would be the magnitude and the direction of the electricforce on a Q= 8.0nC charged particle subsequently place at theorigin?

d) How much work would be required to place Q ate the origin?Assume that the electric potential at infinity is 0.

Thank you!

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 208 Midterm: PHYS 208 TAMU PHYS208 Exam1Spring 16 Solutions

Document Summary

Get access

Related textbook solutions

Physics: Principles with Applications

Related Documents

PHYS 208 Midterm: PHYS 208 TAMU ExamI Spring 2017

PHYSICS 7D Chapter Notes - Chapter 21: Net Force, Atomic Number, Electric Field

PHYS 205 Study Guide - Final Guide: Inductor, Root Mean Square, Electronvolt

Related Questions