For a uniformly charged cylinder of infinite length with radius rr a coaxial cylindrical Gaussian surface (dashed line) of length ll and radius r1<rr1<r is used.

Sunday, 12 August 2018

6:02 PM

Machine generated alternative text: Question: Figure 1. For a uniformly charged cylinder of infinite length witn radius r a coaxial cylindrical Gaussian surface (dashed line) ot length I and radius rl < r is used. A non-conducting cylinder witn a radius T = 2.82 cm nas a uniform charge density, the volume charge density of the cylinder is p = 1.70 C/m . Consider a Gaussian surface inside this non- conducting cylinder. Take the Gaussian surface as a coaxial cylinder with radius rl = 0.829 cm and length I = 83.69 cm. part 1) What is the total charge enclosed by the Gaussian surface? c part 2) What is the electric field strength at a distance Tl = 0.829 cm trom the cylinders axis? part 3) Now considering a Gaussian cylinder witn radius = 8.07 cm and the same length I conducting cylinder. 83.69 cm calculate the electric field strength a distance r2 = 8.07 cm from the centre of the non-

 

Machine generated alternative text: Feedback tor Part 1) The enclosed charge is given by qenc = PV = p x m•21 x l. Remember to convert trom cm to m. Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 Feedback tor Part 2) Figure 1 _ Because the cylinder is infinitely long, the electric field points out from the centre, the electric field vectors E point radially outwards trom the centre of the charged cylinder and are perpendicular to its length. Also the area vectors dA of the Gaussian surface are perpendicular to each surface of the cylinder. Gauss's law tells us that = fi.dA The electric field points out from the centre ot the cylinder (away from positive charge). dA points perpendicularly out from the surface as shown in the figure. You can see that tor the circular ends dA and E are perpendicular, they do not contribute to E • CIAA. Due to symmetry we can assume that the electric field is constant over the surface at "1, this allows us to pull E out the front of the integral so: = = E fdÄ f CIA is the surface area ot the part ottne cylinder at rl, 27TT11 so dA = E • 27T11_ In the first pan we snowed that = pm•211 27rr11 Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11

 

Machine generated alternative text: Feedback tor Part 3) We should use Gauss's law again tor this. In this case the enclosed charge is given by qenc contribute to the total electric flux, so: m•2p. using the same arguments, the circular ends of the cylindrical Gaussian surface do not TRI p 2 wr21Eo Remember to convert from cm to m when you substitute in. Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 A correct answer is 3.07e — 4, which can be typed in as follows: 3.872-4 A correct answer is 796000000, which can be typed in as follows: 796eaaeea A correct answer is 947000000, which can be typed in as follows: 947eaaeea 2r2Eo •

 

 

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