Consider a square with side length l=8.31 mml=8.31 mm placed in an electric field described b

Sunday, 12 August 2018

6:01 PM

Machine generated alternative text: Consider a square with side length I = 8.31 mm placed in an electric tield described by: É (3.41 x 105 + (8.29 x 1010)y2) NIC. where y is in metres. The vertices ottne square are at (0, 0), (0, l), (l, 0) and(l, l) as shown on the diagram. Answer pan 1 in unit vector notation, k is out ottne screen towards you. Figure 1 _ An Electric field in the z-•direction is asymmetric in the x, y- plane. T ne density of electric field lines represents the relative strength of the electric field. In the x direction this is constant but it increasing in the y— directiom part 1) What is the electric field in the centre of the square? part 2) Write a definite integral using dy but not dc that you could solve to calculate the electric flux through this square. = f < spanstyle Where: f < spanstyle —webWt —webkit — tap — highlight — color : (y)dy. Nm2/C. tap highlight — color : rgba(0, 0, 0, 0); " > (y)

 

Machine generated alternative text: and < $1. part 3) Solve this integral to give the flux through the square.

 

Machine generated alternative text: Feedback tor Part 1) You have correctly entered the first component You have correctly entered the second component You have incorrectly entered the third component You need to calculate E at = 5, and y = . In this case, 3.41 x 105 +8.29 x 1010 x {)2) NIC. Remember to convert I from mm to m before substituting in. Marks for this submissiom 0.22/0.33. This submission attracted a penalty of 0.11 Feedback tor Part 2) You have correctly entered the lower limit. You have incorrectly entered the upper limit. You have incorrectly entered the tunctiom The electric flux is given by: In this case, the electric field is constant in the x direction, so fdx So we need to integrate this between y = 0 and y = l. So = = f f Edcdy. l, the length ottne square. Thus: = 1 f Edy. R I x (3.41 x 105 + (8.29 x Marks for this submissiom 0.11/0.33. This submission attracted a penalty of 0.11 Feedback tor Part 3) You need to solve the integral expression from pan 2. —3 8.31X10 3 = 8.31 X 10 (3.41 x 105 + 8.29 x 1010y2)dy. 8.31 XIO — 8.31 X 10-3 3.41 X 105y+ 8.29 X 1010 Y 3 (8.31x10 3 8.31 x 10-3 x 3.41 x 105 x 8.31 x 10-3 +8.29 x 1010 x 3 Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11

 

Machine generated alternative text: A correct answer is 0, which can be typed in as follows: A correct answer is 0, which can be typed in as follows: A correct answer is 1770000, which can be typed in as follows: 177eaae A correct answer is 0, which can be typed in as follows: A correct answer is 0.00831, which can be typed in as follows e. oe831 A correct answer is 689000000 • Y2 + 2830, which can be typed in as follows: 68geaaeea*yA2+283a A correct answer is 155, which can be typed in as follows: 155

 

 

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