Question

0.015 Chapter 19 - Electric Potential Energy and Electric Potential 1. Two point charges are of + 7 C and - 4 HC are held at the corners of a rectangle as shown. The lengths of the sides of rectangle are 0.015 mand 0.05 m. Assume that the electric potential is defined to be zero at infinity *7C 1.a) Determine the electric potential at point A. A) - 2.4 x 10% v B) +3.5 x 10v D) +6.0 x 10V C) -6.5 x 10 V E) - 7.5 x 106 V 0.05 4 цас 1.b). What is the potential difference, V-VA between comers B and A? A) - 2.4m 10% V B) -3.2 x 10' C) - 4.6 x 10 V D) -6.3 x 10 E) - 7.3 x 10 V -1

Answer 1

SOLUTION Question (17 Part 1.a) +7 uc A 0.015 m Let q = +Ful a = - 4UC 010sm B - 4re Seperation of point A from a, (, )= 0.015m in Electric potential at A due to q, (vi) = Rq, i. Vi= gxlog x 7x10-6 0.015 = 4200 x 10 = 42X10S v Seperation of point a from a 2 (22) = 0.05m ... Electric potential at A due to a (V2) = R 92 :. Uz = 9X109 x (-4X166) 0.05 =-720 x 103 22 = -712x105 v

i. Net potential at point A (Unet) = V + V₂ :. (Unet) = 420105 - 7.2x105 = 34.8X1SV = 3.48 8106 v .. [Unet) = 3.5x108 v option (B) is Correct. Part 1.b) Seperation of point B from a (ri)= 0.05m in Electric potential at B due to a, (vi)= ka, ivi= 9x109 x 78106 0.05 = 1.260 x 106 v Seperation of point B from qa (rs) = 0.015m i. Electric potential at B due to az (V2) = k az : V₂ = 9x10x (-4X106) r's 0.015 = -2.4x106/ Met potential at point B (Vinet) = vitval = 1.260x106 2.4x106 = - 1. 14x logu : (Vinet) B

Here, (Unet) A VA (Vinet) B VB .. VB- VA = = - 1.24X106 - 3.58106 = - 4.64X106 :.UB-VA = 4.6X106 v option [c] is Correct.