mat-sln-asn-hwk5-spr02

# mat-sln-asn-hwk5-spr02 - Solution for Homework 5 Gauss Law...

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Unformatted text preview: Solution for Homework 5 Gauss Law Solution to Homework Problem 5.1(Electric Flux out of Room) Problem: A golf tube has total charge- . 1 C . Compute the total electric flux out of the lab room containing the charged golf tube and no other net charge. Select One of the Following: (a) This cannot be solved because the room is not spherical. (b) 0 (c)- 7 10 5 Nm 2 / C (d-Answer)- 1 10 4 Nm 2 / C (e) This cannot be calculated because the distance from the charge to the walls of the room is not given. Solution Gauss Law states e = Q enc =- . 1 C 8 . 85 10- 12 C 2 Nm 2 =- 1 10 4 Nm 2 / C Total Points for Problem: 3 Points Solution to Homework Problem 5.2(Component of Volume Charge Field) Problem: A uniform spherical volume charge has volume charge density = 0 . 02 C m 3 . Inside the charged sphere, the electric field has the form vector E = r 3 r Calculate the x component of the electric field at the point P = (1cm , 2cm , 3cm) . Select One of the Following: (a) 28 N C (b) 15 N C (c) 10 N C (d-Answer) 7 . 5 N C (e) N C Solution The displacement vector modulus is r = radicalbig (1cm) 2 + (2cm) 2 + (3cm) 2 = 14cm and the unit vector is r = parenleftbigg 1cm 14cm , 2cm 14cm , 3cm 14cm parenrightbigg = parenleftbigg 1 14 , 2 14 , 3 14 parenrightbigg So the x component of the electric field at P is vector E P = r 3 r x = (0 . 02 10- 6 C m 3 )( 14 10- 2 m) 3(8 . 85 10- 12 C 2 Nm 2 ) parenleftbigg 1 14 x parenrightbigg = 7 . 5 N C x 1 Total Points for Problem: 3 Points Solution to Homework Problem 5.3(Electric Field Outside a Charged Balloon) Problem: A spherical balloon has a surface charge density of on its outer surface and has radius a . What is the electric field outside the balloon at all points in space? Select One of the Following: (a) r (b) a r r (c-Answer) a 2 r 2 r (d) 4 r 2 r (e) Solution The total charge on the balloon is 4 a 2 . The electric field outside the balloon is vector E ( vector r ) = Q enc 4 r 2 r = 4 a 2 4 r 2 r Total Points for Problem: 3 Points Solution to Homework Problem 5.4(Using Gauss Law to Locate Charge in a Field Map) Problem: Given the field map below, where the field in the regions labelled A and B is unknown, use Gauss Law to determine the sign of the charge in the regions. Select One of the Following: (a) A is positive and B is positive. (b-Answer) A is positive and B is negative. (c) A is negative and B is positive. (d) A is negative and B is negative. A B Solution Gauss Law states the flux leaving the surface is proportional to the charge enclosed. Flux leaving a surface should be approximately proportional to the number of field lines leaving a surface. Since lines leave surface A, a net flux leaves surface A , so by Gauss Law a net positive charge is in surface A . Since lines enter surface B , a net negative flux leaves surface B and by Gauss Law surface...
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mat-sln-asn-hwk5-spr02 - Solution for Homework 5 Gauss Law...

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