Work of Electric Field Problem, can't find mistake

In summary, the person is asking for help with a physics problem involving calculating electric potential and work done in assembling charges. They have already figured out parts A, B, and C, but are struggling with part D. Another person suggests using the formula V= kQ/R for each point in relation to P and then adding them together to solve for W=qV. The total work is calculated to be 2.97*10^-2 J.
  • #1
jigs90
19
0

Homework Statement



I posted the question on cramster at
http://answerboard.cramster.com/physics- ...
so you could see the picture.



Homework Equations



W = qV = kq1q2 / r

The Attempt at a Solution



I figure out A, B, and C, (which are the x component, y component, and the electric potential, but I can't figure out how to do part D, (which is the work to asseble the four charges). Please help! I tried to you W=qV, but that didn't work.

My Answers were
A) was 24.4
B) was 77.96
C) was 2.719x10^3

Work done in assembling +6 uF and -6uF is zero

Total work =kq1q2/r=9*10^9*8.2*16.4*10^-12/(47sin(60

Total work =2.97*10^-2 J
 
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  • #2
The link was

http://answerboard.cramster.com/physics-topic-5-189871-0.aspx

sorry
 
Last edited by a moderator:
  • #3
Wait, should I calculate each PE separately and then add them together? Would there be 6 different PEs?
 
  • #4
Figured it out...
 
  • #5
Upon brief inspection, What I would do is use the formula V= kQ/R for each point in relation to P. Then you can add these based on the sign of the charge to solve for W=qV. Give that a try.
 

1. What is an electric field and how does it work?

The electric field is a region around a charged object where a force is exerted on other charged objects. It is created by the presence of electric charges. The direction of the electric field is defined as the direction a positive test charge would move if placed in the field. The strength of the electric field is determined by the magnitude and distance of the charges.

2. What are some common mistakes made when working with electric fields?

One common mistake when working with electric fields is forgetting to take into account the direction of the electric field. Another mistake is using the wrong units or forgetting to convert units when necessary. Additionally, it is important to consider the charges of all objects involved, not just the ones being directly studied.

3. How do you find the mistake in a problem involving electric fields?

The best way to find a mistake in a problem involving electric fields is to carefully review all the steps and calculations made. Double check all units and conversions, and make sure to consider the direction of the electric field. It can also be helpful to ask a colleague or teacher for assistance in identifying the mistake.

4. Can you give an example of a problem involving electric fields?

Sure! An example problem may involve calculating the electric field at a point due to two charges. The first charge is +3μC located 2 meters to the left of the point, and the second charge is -4μC located 3 meters to the right of the point. By using the formula for electric field (E = kQ/r^2), we can calculate the electric field at the point by first finding the individual electric fields due to each charge, and then adding them together.

5. What are some real-world applications of understanding electric fields?

Understanding electric fields is crucial in many fields, including engineering, physics, and electronics. It is used in the design and operation of electronic devices, such as computers and smartphones. It is also important in the study of electricity and magnetism, which has many practical applications such as generating electricity, powering motors, and lighting homes. Additionally, understanding electric fields is necessary for studying and developing technologies such as wireless charging and electromagnetic levitation.

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