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Unformatted text preview: pan (zp695) – midterm 02 review – chiu – (56565) 1 This printout should have 27 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 0.0 points Throughout a particular region of space, the electric potential can be expressed as V = (4) xz + (2) y + ( − 3) z , what is the magnitude of the electric field at point p = (8 , − 8 , 1) m. Correct answer: 29 . 3428 N / C. Explanation: Recalling that vector E = −∇ V , we calculate vector E , vector E = − parenleftBig (4) z ˆ x + (2)ˆ y + bracketleftBig (4) x + ( − 3) bracketrightBig ˆ z parenrightBig . The magnitude of vector E at p is  vector E  = radicalBig (4) 2 z 2 + (2) 2 + [(4) x + ( − 3)] 2 = radicalBig (4) 2 · (1) 2 + (2) 2 + [(4) · (8) + ( − 3)] 2 = 29 . 3428 N / C . 002 0.0 points The figure shows three very large metal disks carrying the indicated charges. On each surface, the charges are distributed uniformly. Each disk has a very large radius R and small thickness t . The distances between disks are a and b , both much smaller than R . Calculate V 2 − V 1 . 1. 1 2 ǫ πR 2 ( Q 1 a − Q 2 b ) 2. 2 ǫ πR 2 ( Q 2 b − ( Q 2 − Q 1 ) t − Q 1 a ) 3. 1 ǫ πR 2 ( Q 1 a − Q 2 b ) 4. 2 ǫ πR 2 ( Q 2 b − Q 1 a ) 5. 1 2 ǫ πR 2 ( Q 2 b − Q 1 a ) 6. 2 ǫ πR 2 ( Q 1 a − Q 2 b ) 7. 1 ǫ πR 2 ( Q 2 b − ( Q 2 − Q 1 ) t − Q 1 a ) 8. 1 ǫ πR 2 ( Q 2 b − Q 1 a ) correct Explanation: Recall that V = − integraldisplay f i vector E • d vector l . Applying Gauss’ law to either of the plates in the upper region, we find that vector E in the upper region is vector E = Q 1 ǫ πR 2 ( − ˆ y ) . Likewise, in the region between the lower two plates, vector E = Q 1 ǫ πR 2 (+ˆ y ) . Inside the conductors E = 0, so the inte gration along the path inside the conductors makes no contribution. Since E is uniform between the plates, we may calculate Δ V us ing the scalar product Δ V = − vector E • vector l , in this case summed over two regions. Δ V = − parenleftbigg 1 ǫ πR 2 parenrightbigg parenleftBig [ Q 1 ( − ˆ y ) • a ( − ˆ y )] + [ Q 2 (+ˆ y ) • b ( − ˆ y )] parenrightBig = − 1 ǫ πR 2 ( Q 1 a − Q 2 b ) = 1 ǫ πR 2 ( Q 2 b − Q 1 a ) 003 0.0 points pan (zp695) – midterm 02 review – chiu – (56565) 2 A small metal sphere of radius r = 4 mm has an initial charge q = 6 . 1 μ C. A very long ideal copper wire is attached to this sphere and connected to a very distant, large, un charged metal sphere of radius 63 cm. Cal culate the final charge Q on the large sphere, neglecting the small amount of charge on the wire. Correct answer: 6 . 06151 μ C. Explanation: The key point is to realize that once the spheres are connected by the wire, their sur faces must be at the same potential. Since we are told the large sphere is “very distant,” we make the approximation that the spheres are far enough apart that we may neglect the ef fect of one sphere on the potential of the other...
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This note was uploaded on 02/21/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

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