A charge Q = 5.0 nC is located 0.30m from point A and 0.50m from point B. (a)Wha
ID: 1706983 • Letter: A
Question
A charge Q = 5.0 nC is located 0.30m from point A and 0.50m from point B. (a)What is the potential at point A? (b) What is the potential at B? (c) If a point charge q is moved from A to B while Q is fixed in place, through what potential difference does it move? Does its potential increase or decrease? (d) If q = -1.0 nC, what is the change in potential energyas it movesfrom A to B? Does it increase or decrease? (e) How much work is done by the electrical field due to charge Q as q moves from A to B?Explanation / Answer
Quantity of charge located at point at 0.3 m form A and at point at 0.5 m m form B is = 5 n C = 5 *10^-9 C (a)What is the potential at point A V_A = K Q / R where K is the constant = 9*10^ 9 Q = 5 *10^-9 C R = 0.3 m thus, V_ A = (9*10^ 9 )*(5 *10^-9)/ (0.3) = 150 volt (b) What is the potential at point B V_B = K Q / R where K is the constant = 9*10^ 9 Q = 5 *10^-9 C R = 0.5 m thus, V_ A = (9*10^ 9 )*(5 *10^-9)/ (0.5) = 90 volt (c) potential difference between the points A & B is V = V_A - V_ B = 150 vlot - 90 volt = 60 volt Neither increase nor decreased work done = W = V q = (60 volt) (- 1nC) = 60* 10^- 9 J where K is the constant = 9*10^ 9 Q = 5 *10^-9 C R = 0.5 m thus, V_ A = (9*10^ 9 )*(5 *10^-9)/ (0.5) = 90 volt (c) potential difference between the points A & B is V = V_A - V_ B = 150 vlot - 90 volt = 60 volt Neither increase nor decreased work done = W = V q = (60 volt) (- 1nC) = 60* 10^- 9 JRelated Questions
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