ssm_ch24 - Chapter 24 3 (a) An ampere is a coulomb per...

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Chapter 24 3 (a)Anampe reisacou lombpe rsecond ,so 84 A ¢ h= μ 84 C ¢ h s ³ 3600 s h ´ =3 : 0 £ 10 5 C : (b) The change in potential energy is ¢ U = q ¢ V =(3 : 0 £ 10 5 C)(12 V) = 3 : 6 £ 10 6 J. 5 The electric field produced by an infinite sheet of charge has magnitude E = ¾= 2 ² 0 ,whe re ¾ is the surface charge density. The field is normal to the sheet and is uniform. Place the origin of a coordinate system at the sheet and take the x axis to be parallel to the field and positive in the direction of the field. Then the electric potential is V = V s ¡ Z x 0 Edx = V s ¡ Ex; where V s is the potential at the sheet. The equipotential surfaces are surfaces of constant x ;tha t is, they are planes that are parallel to the plane of charge. If two surfaces are separated by ¢ x then their potentials differ in magnitude by ¢ V = E ¢ x =( ¾= 2 ² 0 ) ¢ x .Thu s ¢ x = 2 ² 0 ¢ V ¾ = 2(8 : 85 £ 10 ¡ 12 C 2 = N ¢ m 2 )(50 V) 0 : 10 £ 10 ¡ 6 C = m 2 =8 : 8 £ 10 ¡ 3 m : 19 (a) The electric potential V at the surface of the drop, the charge q on the drop, and the radius R of the drop are related by V = q= 4 ¼² 0 R . Thus R = q 4 ¼² 0 V = (8 : 99 £ 10 9 N ¢ m 2 = C 2 )(30 £ 10 ¡ 12 C) 500 V =5 : 4 £ 10 ¡ 4 m : (b) After the drops combine the total volume is twice the volume of an original drop, so the radius R 0 of the combined drop is given by ( R 0 ) 3 =2 R 3 and R 0 =2 1 = 3 R . The charge is twice the charge of original drop: q 0 =2 q .Thu s V 0 = 1 4 ¼² 0 q 0 R 0 = 1 4 ¼² 0 2 q 2 1 = 3 R =2 2 = 3 V =2 2 = 3 (500 V) = 790 V : 29 The disk is uniformly charged. This means that when the full disk is present each quadrant contributes equally to the electric potential at P
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Consider a ring of charge with radius r and width dr . Its area is 2 ¼rdr and it contains charge dq =2 ¼¾rdr . All the charge in it is a distance p r 2 + D 2 from P , so the potential it produces at P is dV = 1 4 ¼² 0 2 ¼¾rdr p r 2 + D 2 = ¾rdr 2 ² 0 p r 2 + D 2 : The total potential at P is V = ¾ 2 ² 0 Z R 0 rdr p r 2 + D 2 = ¾ 2 ² 0 p r 2 + D 2 ¯ ¯ ¯ R 0 = ¾ 2 ² 0 h p R 2 + D 2 ¡ D i : The potential V sq at P due to a single quadrant is V sq = V 4 = ¾ 8 ² 0 h p R 2 + D 2 ¡ D i = 7 : 73 £ 10 ¡ 15 C = m 2 8(8 : 85 £ 10 ¡ 12 C 2 = N ¢ m 2 ) h p (0 : 640 m) 2 +(0 : 259 m) 2 ¡ 0 : 259 m i =4 : 71 £ 10 ¡ 5 V : 39 Take the negatives of the partial derivatives of the electric potential with respect to the coordinates and evaluate the results for x =3 : 00 m, y = ¡ 2 : 00 m, and z =4 : 00 m. This yields E x = ¡ @V @x = ¡ (2 : 00 V = m 4 ) yz 2 = ¡ (2 : 00 V = m 4 )(( ¡ 2 : 00 m)(4 : 00 m) 2 =64 : 0V = m ; E y = ¡ @V @y = ¡ (2 : 00 V = m 4 ) xz 2 = ¡ (2 : 00 V = m 4 )(3 : 00 m)(4 : 00 m) 2 = ¡ 96 : 0V = m ; E z = ¡ @V @z = ¡ 2(2 : 00 V = m 4 ) xyz = ¡ 2(2
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This note was uploaded on 01/19/2009 for the course PHYS 171.102 taught by Professor Heckman during the Fall '98 term at Johns Hopkins.

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ssm_ch24 - Chapter 24 3 (a) An ampere is a coulomb per...

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