Friday

January 30, 2015

January 30, 2015

Posted by **Aki** on Thursday, September 21, 2006 at 3:38am.

for this, i place the drawing as

A B C

and i am assuming that A + C = B, i am not sure if thats correct or not and i just added the 2 different potential. But then the Vb and 2Vb isnt used and I am not sure where its used unless its telling me that the distance between A and B is half of B and C.

2) Four identical charges (+1.6 ìC each) are brought from infinity and fixed to a straight line. Each charge is 0.36 m from the next. Determine the electric potential energy of this group.

i placed the charges as follow

q1 q2 q 3 q4

and used q1 as the starting point so EPE = 0

then i used the equation V = Kq/r and multiple the V by (1.6 micro columb) to get the EPE. I did that for q2 q3 q4 and changing r as i get closer to q4. Then i added all the EPE together but i am missing something.

EPE of q1 = 0

EPE of q2

(q1)(k)/(.36m) = V

EPE2 = (V)(1.6 micro columb)

EPE of q3

(q1)(k)/(.36m x 2) = V

EPE3 = (V)(1.6 micro columb)

EPE of q4

(q1)(k)/(.36m x 4) = V

EPE4 = (V)(1.6 micro columb)

Add EPE (1 + 2 + 3 + 4) = Total EPE of the group

but neither is correct i am doing stuff wrong, please help. thank you

2) Your EPE of particles 1 and 2 are OK. For particle 3, you need to compute the work needed to approach BOTH the q1 and q2 particles already present. For q4, you need to consider the interaction with q1, q2 and q3. The separation distances will differ in the last two cases.

i change the equation and got thsi

i placed the charges as follow

q1 q2 q 3 q4

and used q1 as the starting point so EPE = 0

then i used the equation V = Kq/r and multiple the V by (1.6 micro columb) to get the EPE. I did that for q2 q3 q4 and changing r as i get closer to q4. Then i added all the EPE together but i am missing something.

EPE of q1 = 0

EPE of q2

(q1)(k)/(.36m) = V

EPE2 = (V)(1.6 micro columb)

EPE of q3

(q)(k)/(.36m) + (q)(k)/(.36m x 2) = V

EPE3 = (V)(1.6 micro columb)

EPE of q4

(q)(k)/(.36m) + (q)(k)/(.36m x 2) + (q)(k)/(.36m x 4) = V

EPE4 = (V)(1.6 micro columb)

Add EPE (1 + 2 + 3 + 4) = Total EPE of the group

**Answer this Question**

**Related Questions**

Physics please clarify - The potential at location A is 379 V. A positively ...

Physics - The potential at location A is 370 V. A positively charged particle is...

Physics - The potential at location A is 379 V. A positively charged particle is...

Physics - Hi, i need help on the following questions. Thank you. 1) The ...

Physics - Hi, i need help on the following questions. Thank you. 1) The ...

Physics - A 245 g particle is released from rest at point A inside a smooth ...

physics - Point A is at a potential of +250 V, and point B is at a potential of...

physics - Point A is at a potential of +220 V, and point B is at a potential of...

physics - The physics of electric potential energy at one corner of a triangle. ...

Physics - Two spherical conductors, A and B , are placed in vacuum. A has a ...