Question

8. DETAILS MY NOTES ASK Two charges are attached to the vertical axis as shown in the figure below. Om (a) Calculate the electric potential at Point A due to the two charges. (b) Calculate the electric potential at Point B due to the two charges. (c) Calculate the electric potential difference of Point A relative to Point B.

Answer 1

qi A F 0.3mm 0.4m q, = +4uG q=-446 B 1 K0Bm D placed 0.09+ 0.16 at C and respectively Let the charges q, and a are Distance between point B and C can be calculated by Pythogonas theorem as (BC)2 = (BD) +(CD)2 BC = 1 (0.3)2+(0.4) - Jo.09+ BC = 0,5m Also, AD = BC=0.5 m da, Distance of change qı fum A = 0,3m da, distance of charge ar from A=0.5m Totul electric potential at A due to two charges, VA Potential due to q VA = t qit 4nto di ( 4x100 hoito 4nt, I de da Potential due toq, t ㅗ 와르 4tto de But, q + 2

0.3 0.5 0:3 0.5 प (6 Electric potential difference of point A relative VA = (9x109) (4x100 + (-4x10%) VA= (9x109) (4x106) / 1 -/ = (36x103) 3:333 – 2 VA - 47,988 x103 4.799 x 10 Volte (Answer) (a IVA= 4.799 x 104 v (by Similarly, for point B Vio a & Az 4ttol (0.5) (0:3) (48100) - 144100 4tto 0.5 - [qx10°) (4x1009) (en + (qx10°) (48100) (-)-333) - 4.799 x 104 Volts Vo = -4.799 x 104 v Ans. Vo = 3) 0.3 0.5 0.3 - to point B is VA-VB = 4.799x10^= (-4.799x104) 9.598x104 v (ans)