Chapter24 - 1. If the electric potential is zero at...

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1. If the electric potential is zero at infinity then at the surface of a uniformly charged sphere it is V = q /4 πε 0 R , where q is the charge on the sphere and R is the sphere radius. Thus q = 4 πε 0 RV and the number of electrons is ( )( ) () ( ) 6 5 92 2 1 9 1.0 10 m 400V 4 2.8 10 . 8.99 10 N m C 1.60 10 C qR V n ee ε 0 × π == = ×⋅ ×
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2. The magnitude is U = e V = 1.2 × 10 9 eV = 1.2 GeV.
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3. (a) An Ampere is a Coulomb per second, so 84 84 3600 30 10 5 Ah Ch s s h C ⋅= F H G I K J F H G I K J .. (b) The change in potential energy is U = q V = (3.0 × 10 5 C)(12 V) = 3.6 × 10 6 J.
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4. (a) V B – V A = U / q = – W /(– e ) = – (3.94 × 10 –19 J)/(–1.60 × 10 –19 C) = 2.46 V. (b) V C – V A = V B – V A = 2.46 V. (c) V C – V B = 0 (Since C and B are on the same equipotential line).
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5. The electric field produced by an infinite sheet of charge has magnitude E = σ /2 ε 0 , where 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 VV E d xV E x s x s =− z 0 , where V s is the potential at the sheet. The equipotential surfaces are surfaces of constant x ; that 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 = ( /2 ε 0 ) x . Thus, x V == ×⋅ × 2 2 885 10 50 010 10 88 10 0 12 2 6 3 ε . . .. CNm V Cm m 2 2 ch bg
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6. (a) () ( ) 15 19 4 4 3.9 10 N 1.60 10 C 2.4 10 N C 2.4 10 V/m. EF e −− = = × × (b) ∆∆ VE s == × = × 24 10 012 29 10 43 .. . . NC m V c h bg
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7. (a) The work done by the electric field is 19 12 2 00 0 12 2 2 0 21 (1.60 10 C)(5.80 10 C/m )(0.0356 m) 2 2 2(8.85 10 C /N m ) 1.87 10 J. fd i qq d Wq E d s d z σσ εε −− ×× =⋅ = = = ×⋅ ∫∫ G G (b) Since V – V 0 = – W / q 0 = – σ z /2 ε 0 , with V 0 set to be zero on the sheet, the electric potential at P is 12 2 2 12 2 2 0 (5.80 10 C/m )(0.0356 m) 1.17 10 V. 22 ( 8 . 8 5 1 0 C / N m ) z V ε × =− ×
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8. (a) By Eq. 24-18, the change in potential is the negative of the “area” under the curve. Thus, using the area-of-a-triangle formula, we have VE d s x −= ⋅= = z 10 1 2 22 0 0 2 G G bg b g which yields V = 30 V. (b) For any region within 0 3 << − ⋅ z xE d s m, G G is positive, but for any region for which x > 3 m it is negative. Therefore, V = V max occurs at x = 3 m. d s x = z 10 1 2 32 0 0 3 G G which yields V max = 40 V. (c) In view of our result in part (b), we see that now (to find V = 0) we are looking for some X > 3 m such that the “area” from x = 3 m to x = X is 40 V. Using the formula for a triangle (3 < x < 4) and a rectangle (4 < x < X ), we require 1 2 1 20 4 20 40 bgb g b gb g +− = X . Therefore, X = 5.5 m.
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9. We connect A to the origin with a line along the y axis, along which there is no change of potential (Eq. 24-18: G G Ed s ⋅= z 0). Then, we connect the origin to B with a line along the x axis, along which the change in potential is VE d s x d x x =− F H G I K J z z = G G 400 4 2 2 0 4 0 4 ..
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter24 - 1. If the electric potential is zero at...

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