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Answer to Question #6254 in Electric Circuits for Tracy

Question #6254
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
Expert's answer
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

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