The two charges in the figure below are separated by a distance d = 3.00 cm, and
ID: 1265785 • Letter: T
Question
The two charges in the figure below are separated by a distance d = 3.00 cm, and Q = +6.70 nC.
(a) Find the electric potential at A.
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. kV
(b) Find the electric potential at B.
kV
(c) Find the electric potential difference between B and A.
V
Explanation / Answer
Solution :
The electric potential at the point A due to Q = Va =kQ/d = (9x10^9) (6.7x10^-9) /(3x10^-2)
= 20.1 x 10^2 volts
Electric potential at A due to 2Q = k (2Q)/ sqrt(d^2+d^2) =(9x10^9)(2X6.7x10^-9)/ (1.414 x3x10^-2)
= 28.4 x 10^2 volts
total electric potential at A due to the two charges = VA = 48.5 x10^2 =4.85 kV
b) distance between Q and the point B =sqrt(d^2+d^2) =1.414x d x10^-2 m
Electric potential at B due to Q = (9x10^9)(6.7x10^-9)/(1.414 x 3 x10^-2) =14.2 x 10^2 V
electric potentil due to 2Q =(9x10^9)(2x6.7x10^-9)/(3x10^-2)
= 40.2 x10^2 volts
Total electric potential at B due to Q and 2Q =(14.2+40.2 )x10^2
= 54.4 x10^2 volts
= 5.44 kV
c) potential difference between the two points = 5.44 - 4.85
= 0.59 kilovolts
= 591.28 volts
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