Physics 2702 Exam2, Spring-2015 Name: 18) The point charges, ah and -42each inc
ID: 1422248 • Letter: P
Question
Physics 2702 Exam2, Spring-2015 Name: 18) The point charges, ah and -42each inc (10° C) are placed as shown in the figure. Each square in the graph is 1 cm (0.01m), there are six points in the graph. The goal is to calculate the voltages at each point and figure out which location will give a highest VE KO and lowest voltage value among all the six locations. a) What are the voltages at a, and f? V, X 10" (1x10 ). 30 0 0 3rd a 9x 10 kids b b) What are the voltages at b and e? V a X10 (IX 10 " . v. 9 x 10' (1x10 ) 06 D3 ... are the voltages at cand d?Explanation / Answer
a) Va = (9*10^9*1*10^-9*10^2)*[1/2 - 1/6] = 300 V
Vf = (9*10^9*1*10^-9*10^2)*[1/6 - 1/2] = -300 V
b) Vb = (9*10^9*1*10^-9*10^2)*[1/1 - 1/5] = 720 V
Ve = (9*10^9*1*10^-9*10^2)*[1/5 - 1/1] = -720 V
c) Vc = Vd = 0 as distance from both charges is same and charges are of opposite sign.
d) Vb = Highest = 720 V
Ve = lowest = -720 V