Resultant electric field magnitude and direction

In summary, the problem involves a square with four charges at each corner and the task is to determine the resultant electric field at the center of the square, as well as the force on a point charge placed at that point. The superposition of electric fields is used to solve the problem, by adding the individual electric fields as vectors. After solving, the magnitude of the net field is found to be 8000 Cm^-1 at an angle of 18.43°, and the force on the point charge is calculated to be 4.8*10^-5N.
  • #1
ledwardz
7
0

Homework Statement



a square of sides 10cm has four charges at each corner starting clockwise from top right they are 5,10,6 and 3nC. Determine the resultant electric field including magnitude and direction at the centre of the square. Owww and the medium is air. Finally what is the force on a 6pC placed at this point.

i should probably point out I've never done physics in my life be4 u give me grief... take pity


Homework Equations



okay first part I am using

kQ/R^2
k= 1/4pi* E0*Er
E0= 8.854*10^-12
Er= 1.005
my value of r is 7.07cm which = 0.0707m

sooooo i have 4 electric fields of;

8945.723 Cm^-1
17892.28 ...
10735.37 ...
5367.68 ...

solving as vectors...

(x)= 5367.68cos45 + 10735.37cos45 - 17892.28cos45 - 8945.723cos45
= 7590.55 to the left

(y)= 10735.37sin45+17892.28sin45 - 8945.723sin45 - 5367.68sin45
= 2530 upwards

gives me a magnitude of 8000 Cm^-1
at an angle of 18.43

then the force is

q*E = 6*10^-9 *8000 = 4.8*10^-5N

Thanks for any help, Lee.:cry:
 
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  • #2
use superposition of electric fields, adding them as vectors. you will find the net field's magnitude as well as direction easily.
 
  • #3
supratim1 said:
use superposition of electric fields, adding them as vectors. you will find the net field's magnitude as well as direction easily.



I think that is what i did, have i got the right idea?
 
  • #4
ledwardz said:
...

solving as vectors...

(x)= 5367.68cos45 + 10735.37cos45 - 17892.28cos45 - 8945.723cos45
= 7590.55 to the left  I agree.

(y)= 10735.37sin45+17892.28sin45 - 8945.723sin45 - 5367.68sin45
= 2530 upwards  I think you missed adding the 10735.37sin45°. Otherwise it looks good.

Re-Do what follows.


gives me a magnitude of 8000 Cm^-1
at an angle of 18.43

then the force is

q*E = 6*10^-9 *8000 = 4.8*10^-5N

Thanks for any help, Lee.:cry:

supratim1 said:
use superposition of electric fields, adding them as vectors. you will find the net field's magnitude as well as direction easily.
Hello Lee.

See my comments in red above.

Your method looks good. You appear to have merely missed one term in the y component!
 
  • #5
SammyS said:
Hello Lee.

See my comments in red above.

Your method looks good. You appear to have merely missed one term in the y component!



cheers man, appreciate the help. good to know i am doing summit right for once:approve:e.
 

Related to Resultant electric field magnitude and direction

1. What is a resultant electric field?

The resultant electric field is the overall electric field at a given point in space, taking into account the contributions of all surrounding charges.

2. How is the magnitude of a resultant electric field calculated?

The magnitude of a resultant electric field is calculated by vector addition of the individual electric fields from each charge at the given point. This can be done using the Pythagorean theorem and trigonometric functions.

3. What factors affect the magnitude of a resultant electric field?

The magnitude of a resultant electric field is affected by the strength and distance of the individual charges contributing to the field, as well as the medium through which the electric field is passing.

4. How is the direction of a resultant electric field determined?

The direction of a resultant electric field is determined by the direction of the individual electric fields at the given point, as well as their relative strengths and angles of inclination.

5. What is the significance of the direction of a resultant electric field?

The direction of a resultant electric field indicates the direction that a positive test charge would move if placed at that point. This helps us understand the behavior and interactions of charges in an electric field.

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