workout 25a

# workout 25a - HPHY 253 Workout 25a The Electric Potential...

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Unformatted text preview: HPHY 253 Workout 25a The Electric Potential Dr. A. E. Bak Name 1. Serway & Jewett Problem 25 : 13 . At a certain distance from a charged particle, the magnitude of the electric ¡eld is 500 V =m and the electric potential is & 3 : 00 k V . (a) What is the distance to the particle? For the point charge in question, we have 500 = E = k e j q j r 2 ; & 3000 = V = k e q r : It follows that r = & & & & V E & & & & = 6 : 00 m is the distance to the particle. (b) What is the charge involved here? On making use of our expression for the electric potential, we &nd that q = V r k e = ( & 3000) (6 : 00) 8 : 987551787 ¡ 10 9 = & 2 : 00 ¡ 10 & 6 C = & 2 : 00 & C is the charge involved here. 2. From Serway & Jewett Problem 25 : 61 . An electric dipole is located along the y axis as shown in Figure P 25 : 61 . The magnitude of its electric dipole moment is p = 2 Qa . At a point P , which is far from the dipole (that is, r ¢ a ), show that the electric potential is given by V £ k e p (cos ¡ ) =r 2 . Consider the two triangles shown in the &gure. On making use of the law of cosines, we &nd that r 2 1 = r 2 + a 2 & 2 ra cos ¡ ; r 2 2 = r 2 + a 2 & 2 ra cos ( ¢ & ¡ ) = r 2 + a 2 + 2 ra cos ¡ : Note that r in this problem denotes the distance from the origin. The electric potential at P due to the positive charge is given by V 1 = k e (+ Q ) r 1 = k e Q p r 2 + a 2 & 2 ra cos ¡ = k e Q r ¡ 1 + a 2 r 2 & 2 a cos ¡ r ¢ & 1 = 2 ; and the electric potential at P due to the negative charge is given by V 2 = k e ( & Q ) r 2 = & k e Q p r 2 + a 2 + 2 ra cos ¡ = & k e Q r ¡ 1 + a 2 r 2 + 2 a cos ¡ r ¢ & 1 = 2 : On making use of the principle of superposition, we &nd that the total electric potential at P is given by V = V 1 + V 2 = k e Q r " ¡ 1 + a 2 r 2 & 2 a cos ¡ r ¢ & 1 = 2 & ¡ 1 + a 2 r 2 + 2 a cos ¡ r ¢ & 1 = 2 # : If point P lies very far away from the dipole, then a=r ¤ 1 and ¡ 1 + a 2 r 2 ¥ 2 a cos ¡ r ¢ & 1 = 2 £ 1 & 1 2 ¡ a 2 r 2 ¥ 2 a cos ¡ r ¢ £ 1 ¦ a cos ¡ r ; where we have made use of the binomial approximation (1 + £ ) n £ 1 + n£ for j £ j ¤ 1 and j n£ j ¤ 1 : It follows that V £ k e Q r £¡ 1 + a cos ¡ r ¢ & ¡ 1 & a cos ¡ r ¢¤ = k e Q r ¡ 2 a cos ¡ r ¢ = k e p cos ¡ r 2 ; which is the electric potential at P due to a point dipole of moment...
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## This note was uploaded on 05/01/2011 for the course HPHY 253 taught by Professor Bak during the Spring '09 term at Morehouse.

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workout 25a - HPHY 253 Workout 25a The Electric Potential...

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