PHY 2049 Midterm: University of Florida PHY 2049 exam4sol

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turquoisegnu128

7 Mar 2019

School

University of Florida

Department

Physics

Course

PHY 2049

Professor

All

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Exam 4 solution: four charges are arranged as shown in the gure. Each charge is 8 cm from the origin. The values of the charges are q1 = +3 c, q2 = +1 c, q3 = +1 c, and q4 = 1 c. Solution: the x-component of the force is fx = k(3 c 1 c)(2 c)/r2, and the y-component of the force is fy = k(1 c 1 c)(2 c)/r2, where r = 8cm. The magnitude of the force at the origin is qf 2 x + f 2 y : a wire segment of length l has constant linear charge density > 0. Which of the following expressions gives the magnitude of the electric eld a distance. Solution: this problem is identical to one from exam 1 - see that solution: a point charge lies at the center of two concentric conducting shells, as shown in the gure.

Related Questions

Coulomb's law for the magnitude of the force F between twoparticles with charges Q and Q' separated by a distance, d is |F| =K *|QQ'|/d2 where K=1/4peo = 8.854*10^-12 nC is the permittivity offree space.

Consider two point charges located on the x axis: one charge, q1 =-19.5nC , is located at x1= -1.680m ; the second charge, q2=31.5nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a thirdcharge q3 = 52.5nC placed between q3 and at x3= -1.245m ? Youranswer may be positive or negative, depending on the direction ofthe force.

**Please solve with a numerical answer. I've tried this severaltimes and this is my last attempt on mastering physics.

Coulomb's law for the magnitude of the force between two particleswith charges and separated by a distance is |F| = K |QQ'|/d^2 whereK = 1/4piEo, Eo = 8.854 x 10^-12 C^2/Nxm^2. is the permittivity offree space.

Consider two point charges located on the x axis: one charge, q1 =-11.5nC , is located at x1 = -1.715 m; the second charge, q2 =34.5nC , is at the origin x = 0.00000.

What is the net force exerted by these two charges on a thirdcharge q3 = 49.5nC placed between q1 and q2 at x3 = -1.240m ?

Your answer may be positive or negative, depending on the directionof the force.

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.

PHY 2049 Midterm: University of Florida PHY 2049 exam4sol

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