Net Electric Field Homework: Find Ea, Eb, and Enetx

Click For Summary
SUMMARY

The discussion focuses on calculating the net electric field (Enetx) at a point due to two charges using the formula Enetx=((kq1)/r^2 +^2). The user initially miscalculated the electric field components, resulting in an incorrect total of -308.335 N/C. The correct approach involves using absolute values for charge magnitudes and determining direction afterward, leading to a revised calculation of -241.467 N/C. Participants emphasized the importance of sketching vector directions for clarity in calculations.

PREREQUISITES
  • Understanding of electric field concepts and vector addition
  • Familiarity with Coulomb's Law and its application
  • Knowledge of basic algebra and calculus for physics
  • Experience with electric field calculations in one-dimensional scenarios
NEXT STEPS
  • Study the application of Coulomb's Law in electric field calculations
  • Learn about vector addition in physics, particularly for electric fields
  • Explore the concept of electric field lines and their representation
  • Practice problems involving multiple charges and net electric field calculations
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to improve their understanding of electric field calculations and vector analysis.

Jrlinton
Messages
133
Reaction score
1

Homework Statement


204.PNG


Homework Equations


Enetx=((kq1)/r^2 +^2) with the pos/neg of the coefficients being dependent on the position of the charges around the point

The Attempt at a Solution


As everything is one dimensional we can assume the j an k compnents of the electric field are 0
Part a
Ea= -(kq1)/r1^2+(kq2)/r2^2
=-(8.99E9N)(3.7E-12C)/(.011m)^2+(8.99E9N)(-1.8E-12C)/(.022m)^2
=-274.901 N/C+ -33.4339 N/C
=-308.335 N/C

This was incorrect and I am unsure of my mistake(s) and can only assume I would carry them over to the next two calculations
 
Physics news on Phys.org
Jrlinton said:
-274.901 N/C+ -33.4339 N/C
At A, the positive charge provides a negative component to the left and the negative charge provides a positive component to the right. You show two negative components. In short, you used the negative sign for the negative charge twice. It is probably easier to calculate the size of the component using positive values for all charges, figure out the direction of the field components from the diagram and add negative signs as needed.
 
kuruman said:
At A, the positive charge provides a negative component to the left and the negative charge provides a positive component to the right. You show two negative components. In short, you used the negative sign for the negative charge twice. It is probably easier to calculate the size of the component using positive values for all charges, figure out the direction of the field components from the diagram and add negative signs as needed.

So you're saying I should use absolute values for both charges and hen decide direction when adding them together? So in this case it should have been -274.901+33.4339= -241.467 N/C?
 
The signs of the charges along with the relative locations of the charges at the points of interest are mucking up your algebra. You either have to be very careful indeed to keep it all straight, or do the practical thing: First sketch in vectors for the field directions due to each charge at the points of interest. Then calculate the field magnitudes using absolute values for the charges. Write the algebra to incorporate direction based on the vector directions.Edit: Ah. I see that kuruman got there ahead of me! :smile:
 
Jrlinton said:
So you're saying I should use absolute values for both charges and hen decide direction when adding them together? So in this case it should have been -274.901+33.4339= -241.467 N/C?
That's what I and @gneill are saying. Now you should be able to do the other two parts on your own. If not, you know where to go for help.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
4
Views
2K
Replies
18
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K