Determine the potential energy of the configuration

[Pruddy]

. Homework Statement

Diagram: Please check the attachment. It has a diagram that will help solve the problem

#6)Point charges Q1 and Q2 are located at two vertices of an equilateral triangle. Point P is the third vertex.The length of each side of the triangle is 28.0 cm. Q1 = +26e, and Q2 = -14e
#7). Point charge, Q3 = +7e, is now placed at point P. Determine the potential energy of the configuration formed
by Q1, Q2 and Q3. (Assume that the potential energy is zero at infinity.)

Homework Equations

V= k(q)/r, k= 9.99 x 10^9

The Attempt at a Solution

#6)I found the potential energy the potential energy of both point a and point p. Using the distance r = 0.28cm. Then I sum the potential energy of both points to get the potential energy at P.
#7) I used thesame procedure for the second problem.

I will be very grateful, if anyone can help. Thanks a lot.

 

[haruspex]

Where's point a? Did you mean you found the potential (not potential energy) at P due to each of q1 and q2 separately?
If you want your answers checked pls post all your working.

 

[Pruddy]

I am sorry if wasnt very clear. what i meant was i found the potential of q1 and q2. Took the sum of both potential. Please check the attachment the question is number 7 and 6. Thank you

 

[Pruddy]

Please check the attachment...

 

[haruspex]

i found the potential of q1 and q2. Took the sum of both potential. Please check the attachment the question is number 7 and 6.

Still not very clear, but I assume you mean you found the potential at P due to q1, the potential at P due to q2, and added them. That's fine for the first part.
"Same for the second" is much too vague. Pls post exact working.

 

[Pruddy]

These are my workings for #1
V= kQ/R
K= 9 x 10^9
q1 = 26e
q2 = -14e
e = 1.602 x 10^-19
r = 28 cm

Using V= kQ/r, I found the sum of the potential of P due q1 and q2
So,
(9 x 10^9 x 26 x 1.602 x 10^-19/0.28) + (9.0 x 10^9 x 14 x 1.602 x 10^-19/0.28)

=1.3388 x 10^-7 + 7.209 10^-8

=2.059 x 10^7
But for some reason my answer is wrong.

 

[Pruddy]

Working for #2
V= kQ/R
K= 9 x 10^9
q1 = +26e
q2 = -14e
q3 = +7e
e = 1.602 x 10^-19
r = 28 cm

Using V= kQ/r, I found the electric potential at q1, q2 and q3. then took their sums.
So,
(9 x 10^9 x 26 x 1.602 x 10^-19/0.28) + (9.0 x 10^9 x 14 x 1.602 x 10^-19/0.28) + (9 x 10^9 x 7 x 1.602 x 10^-19/0.28)

=1.3388 x 10^-7 + 7.209 10^-8 + 3.6045 x 10^-8
= 2.420 x 10^-7

 

[Pruddy]

#6)Point charges Q1 and Q2 are located at two vertices of an equilateral triangle. Point P is the third vertex.The length of each side of the triangle is 28.0 cm. Q1 = +26e, and Q2 = -14e
#7). Point charge, Q3 = +7e, is now placed at point P. Determine the potential energy of the configuration formed
by Q1, Q2 and Q3. (Assume that the potential energy is zero at infinity.)

 

[SammyS]

Working for #2
V= kQ/R
K= 9 x 10^9
q1 = +26e
q2 = -14e
q3 = +7e
e = 1.602 x 10^-19
r = 28 cm

Using V= kQ/r, I found the electric potential at q1, q2 and q3. then took their sums.
So,
(9 x 10^9 x 26 x 1.602 x 10^-19/0.28) + (9.0 x 10^9 x 14 x 1.602 x 10^-19/0.28) + (9 x 10^9 x 7 x 1.602 x 10^-19/0.28)

=1.3388 x 10^-7 + 7.209 10^-8 + 3.6045 x 10^-8
= 2.420 x 10^-7

#6)Point charges Q1 and Q2 are located at two vertices of an equilateral triangle. Point P is the third vertex.The length of each side of the triangle is 28.0 cm. Q1 = +26e, and Q2 = -14e
#7). Point charge, Q3 = +7e, is now placed at point P. Determine the potential energy of the configuration formed
by Q1, Q2 and Q3. (Assume that the potential energy is zero at infinity.)

This not the way to approach the second problem, (#7).

Starting with all three charges at infinity:

How much work does it take to bring q₁ to its position in the triangle?

After placing q₁ in position, how much work does it take to bring q₂ to its position in the triangle?

After placing q₁ and q₂ in position, how much work does it take to bring q₃ to its position in the triangle?​

Adding all of those together, what does the result tell you?

 

[Pruddy]

Got It! Thanks a lot.