Above is a cross section of  spherical shell made of conducting material, with a spherical open cavity in the centre.

Sunday, 12 August 2018

4:06 PM

 

 

Machine generated alternative text: Question: Figure 1 _ Above is a cross section ot spherical shell made ot conducting material, witn a spherical open cavity in the centre. The spherical shell has an inner radiusT1 and outer radius "2 with a total charge Q A conducting sphere has a radius r2 = 7.01 cm. The sphere has a total charge Q = +14.82gC. Inside the sphere is a spherical cavity (empty space) witn radius rl = 1.67 cm which encloses a charge q = +3.32"C. part 1) What is the electric field strength inside the conducing sphere at radius = 4.51 cm? Your last answer was interpreted as follows: 0 part 2) What total charge is found on the outer surface at "2? Qr2 = O Your last answer was interpreted as follows: 0 part 3) What is the electric field strength at a distance = 20.71 cm trom the center of the sphere? (Note this is outside the sphere.) Your last answer was interpreted as follows: 0

 

Machine generated alternative text: Your answer is partially correct Note: Your answers should be correct to 3 significant figures. Feedback tor Part 1) You have correctly answered this part Marks for this submissiom 0.33/0.33. Feedback tor Part 2) The total charge on the two surfaces at rl and is given by -HQ as in a conductor the charges are tound on the surface. To give zero field inside the conductor, the charge on the surface at rl must cancel out the -eq charge inside the cavity. This is due to Gauss' law: It E is zero then 0 qenc Thus at there must be a charge ot —q. Total charge on sphere Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 Feedback tor Part 3) Draw a spherical Gaussian surface witn radius r4. Figure 1 _ A spherical Guassian surface (dashed line) of radius > is set up outside the spherical Snell. Gauss's law tells us that In this case the enclosed charge qenc = q + Q. The electric field points out from the sphere (away from positive charge) and dA also points out from the surface so these are parallel. Due to symmetry, E will be constant over the surface so can be pulled out the front of the integral: fÉdA = EfdA = E • = q+Q

 

Machine generated alternative text: fdA is the total surface area ot the Gaussian surface, a sphere with radius r 4 in this case so: E04m•24 • Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 A correct answer is 0, which can be typed in as follows: A correct answer is 18.1, which can be typed in as follows: 18.1 A correct answer is 3800000, which can be typed in as follows: 380eaae

 

 

Untitled picture.png Machine generated alternative text: Question: Figure 1 _ Above IS a cross section ot spherical shell made ot conducting material, witn a spherical open cavity in the centre. The spherical shell has an inner radiusT1 and outer radius "2 with a total charge Q. A conducting sphere has a radius r2 = 4.58 cm. The sphere has a total charge Q = +15.52gC. Inside the sphere is a spherical cavity (empty space) witn radius rl = 1.09 cm which encloses a charge q = +1.18"C. part 1) What is the electric field strength inside the conducing sphere at radius = 2.94 cm? Your last answer was interpreted as follows: 0 part 2) What total charge is found on the outer surface at "2? Qr2 = 16.7 Your last answer was interpreted as follows: 16.7 part 3) What is the electric field strength at a distance = E = 8310000 13.52 cm trom the center of the sphere? (Note this is outside the sphere.) Your last answer was interpreted as follows: 8310000 Untitled picture.png Machine generated alternative text: Correct answer, well done. Feedback tor Part 1) You have correctly answered this part Marks for this submissiom 0.33/0.33. Feedback tor Part 2) You have correctly answered this part Marks for this submissiom 0.33/0.33. Feedback tor Part 3) You have correctly answered this part Marks for this submissiom 0.33/0.33. Ink Drawings Ink Drawings Ink Drawings         
Untitled picture.png Machine generated alternative text: Correct answer, well done. Feedback tor Part 1) You have correctly answered this part Marks for this submissiom 0.33/0.33. Feedback tor Part 2) You have correctly answered this part Marks for this submissiom 0.33/0.33. Feedback tor Part 3) You have correctly answered this part Marks for this submissiom 0.33/0.33. Ink Drawings  

 

 

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