# Answer on Electric Circuits Question 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

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

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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