Answer to Question #6254 in Electric Circuits for Tracy
Point A is at position (3j -4k)cm. Point B is at position -4cm k. Point C is at position (3i -2j + 5k) cm. Point D is at position 3cm j. Charges of +4 and -9uC are at A and B respectively. What is the electric potential energy of an electron at D?
e=1.6x10^-19C and k=9x10^9 Nm^2/C^2
Since point C has no charge, we can ignore it. The electric potential energy E(D;A,B) at point D in the electric field generated by charges A and B is equal to the sum E(D;A,B) = E(D;A) + E(D;B), where E(D;A), resp. E(D,B), is the electric potential energy of electron at D in the electric field generated by A, resp.
It is known that E(D;A) = k e Q(A) / AD, where Q(A)=+4uC = 4x10^-6 C is the charge of A, and AD is the distance between A and D. Similarly, E(D;B) = k e Q(B) / BD.
Let us write down coordinates of points A,B,D:
A(0,3,-4), B(0,0,-4), D(0,3,0)
Then AD = square_root( (3-3)^2 + (0+4)^2 ) = 4 cm = 0.04 m
BD = square_root( (3-0)^2 + (0+4)^2 ) = = square_root( 9+16 ) = 5 cm = 0.05 m
Hence E(D;A,B) = E(D;A) + E(D;B) = = k e Q(A) / AD + k e Q(B) / BD = = k e ( Q(A) / AD + Q(B) / BD ) = = 9*10^9 (-1.6*10^-19) ( 4*10^-6 / 0.04 - 9*10^-6 / 0.05 ) = = -9 * 1.6 * 10^-9 * 10^-4 ( 4/4 - 9/5 ) = = -9 * 1.6 * 10^-13 * (-0.8) = = 1.52 x 10^-12 J