Electric Field Calculations for Stationary Charges and a Dipole at Point P(0,4m)

Click For Summary
SUMMARY

The discussion focuses on calculating the electric field vector at point P(0,4m) due to two stationary charges, q1 = +2mC and q2 = -15mC, and determining the torque on a dipole located at the same point. The electric field equations used are E = k*q/r², with calculated values E1 = 4.5 x 10^6 N/C and E2 = 1.7 x 10^6 N/C. Clarifications were made regarding the vector nature of electric fields, emphasizing that while components can be negative, the vector itself is not described as such. Additionally, guidance was provided on correctly determining angles using trigonometric functions.

PREREQUISITES
  • Understanding of electric field concepts and vector quantities
  • Familiarity with Coulomb's law and the equation E = k*q/r²
  • Basic knowledge of trigonometry for angle calculations
  • Experience with significant figures in scientific calculations
NEXT STEPS
  • Study the principles of electric fields and forces in electrostatics
  • Learn about torque on dipoles in electric fields
  • Explore vector addition and decomposition in physics
  • Review significant figures and their importance in scientific measurements
USEFUL FOR

Students in physics, particularly those studying electrostatics, as well as educators and anyone involved in teaching or learning about electric fields and dipole interactions.

BHH
Messages
1
Reaction score
0

Homework Statement


The picture below shows two stationary charges q1= +2mC and q2= -15 mC.
and I need to find the electric field vector at field point P(0,4m). E= [px, py] .

Also
A little dipole is positions in P(0,4m) and is given with p= [px,py]=[2*10^-8 ,0] Cm.

Find the torque that acts on the dipole because of the electric field and its potential energy

Homework Equations


E=k*q/r^2

The Attempt at a Solution


I have found out that E1=4.5*10[6][/SUP] and E2= 1.7*10^6. I am not sure if the electric field E1 is negative or positive ? I think its negative because the vector goes downwards. And when I find the angle for E2 do I use arctan(8/2)?
 

Attachments

  • Screen Shot 2016-01-14 at 16.24.40.png
    Screen Shot 2016-01-14 at 16.24.40.png
    2.5 KB · Views: 476
Physics news on Phys.org
WELCOME TO PF.
BHH said:
I have found out that E1=4.5*10[6][/SUP] and E2= 1.7*10^6.
OK. But I would keep an extra significant figure while doing the calculations. Then round off to the appropriate number of significant figures at the end. Also, don't forget the units.
I am not sure if the electric field E1 is negative or positive ? I think its negative because the vector goes downwards.
Electric field is a vector quantity. It is not good language to say that a vector is negative. However, the components of a vector can be negative or positive. So, if the field points downward (along the negative y axis) then you would say that the y component of the electric field is negative.

And when I find the angle for E2 do I use arctan(8/2)?
This does not look right. Are you trying to find the angle that E2 makes to the horizontal direction or the vertical direction? If you are constructing a right triangle to help find the angle, make sure you have the dimensions of the sides of the triangle correct.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
1K
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K