(20%) Problem 3: A point charge Q = 3.4 nC is located at the origin. A cylinder

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(20%) Problem 3: A point charge Q = 3.4 nC is located at the origin. A cylinder of radius R-045 m is concentric with the z-axis with the top of the cylinder at height H1 = 0.75 m above the origin and the bottom of the cylinder at height H2 = 0.35 m below the origin. Refer to the figure. The y-axis points into the screen z-axis x-axis Otheexpertta.com 20% Part (a) Calculate the electric flux passing out through the top surface in units of N·m2/C. Grade Summary Deductions Potential 0% 100% cosO sin) cotan0asin0 atan) acotan) Submissions Attempts remaining: 3 (4% per attempt) detailed view tan acosO tanh)cotanh() Degrees Radians Submit Hint I give up! Hints: 4% deduction per hint. Hints remaining: 3 Feedback: 5% deduction per feedback 20% Part (b) Calculate the electric flux passing out through the bottom surface in units of N·m2/C. 20% Part (c) Calculate the electric flux passing out through the side surface in units of N-m2/C 20% Part (d) Sum the fluxes passing out through all three surfaces of the cylinder to determine the total flux from the cylinder in units of N·m2/C. 20% Part (e) Use Gauss's Law to determine the total flux from the cylinder in N·m2/C

Explanation / Answer

A) Electri flux is the electric field per unit area.

Electric field at top surface can be fund using coulomb's law

E= K q/r^2

where K - coulombs const= 8.99* 10^9 N.M^2/C^-2

q given = 3.4 nC= 3.48 10^-9 C

r is the distance from the charge to the point where electric field is measured.

Here r = H1 =.35m

First we will calculate electric field.

So using eqn 1, E = (8.99*10^9*3.4*10^-9)/ (.75)^2 -----------(EVERY THING IS IN SI UNITS)

E = 54.34 N/C ( UNIT YOU CAN PUT FOR EACH AND FIND OUT OR JUST REMEMBER "E"

IS THE FORCE PER UNIT CHERGE SO FORCE UNIT IS N(NEWTON/ CHARGEUNIT IS coULOMB)

now Flux is nothing but electric field per unit area.

So we need to calculate the area of the surface.

Area of top surface A= Pi * R^2(Area of circle) given R -.45 m

A= 3.14* (.45)^2 m^2

A=.636 m^2

So flux1(Phi1) = E/A1

Flux1 = 54.34/.636 = 85.44 N/(C.m^2)

B) In the same way you can find out the Electric filed at bottom surface and is given by

E2 = 8.99*10^9*3.4*10^9 /(.35)^2

E2 = 249.52 N/C

Area of the bottom surface= Area of the top surface = .636 m^2

Flux_2 = E2/ A = 249.52/.636 = 392.324 N/(C.m^2)

C) Electric flux passing throgh the side

distance from the charge to sides of the cylinder is Radius of the cylinder = .45m

So using eqn 1 Electric field at side is given by E3 = 8.99*10^9*3.4*10^-9/(.45)^2

E3 = 150.9 N/C

Flux_3 = E3/ Area of the sides

Area of the side of the cyrinder = 2* Pi* r* (total height of the cylinder)

r = .45m

h= h1+ h2

h= .35+.75 = 1.1

Area of the side = 2*3.14* .45* 1.1 =3.1086

Flux_3 = E3/ Area = 150.9/3.1086

Flux_3 = 48.542 N/(C.m^2)

d) total flux = flux 1+ flux 2+ flux 3 = 85.44+392.324+48.542= 526.306 N/(C.m^2)

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