What is the Strength of the Electric Field?

AI Thread Summary
The discussion focuses on calculating the strength of the electric field using the formula E1 = (k*p)/(r^2 + a^2)^(3/2). The initial attempt incorrectly used the charge instead of the dipole moment for p, which should be calculated as p = 2aq. After correcting this, a new calculation yields a result of 906, raising concerns about the expected relationship between the electric field strength and the distance r. The complete problem statement emphasizes the need to express the answer with two significant figures. Overall, the thread highlights common pitfalls in applying formulas related to electric fields and dipole moments.
littlejon
Messages
20
Reaction score
0

Homework Statement


Specify the strength of the electric field. Let r = 7.3cm .

Homework Equations


E1= (k*p)/(r^2+a^2)^3/2
Here k= constant used in Coulomb's law. a= half the length of dipole
p= dipole moment = charge x length of dipole. Charge = 3nC and length of dipole

The Attempt at a Solution


E1= (9E9)(3E-9)/(.146)^2+(.073)^2)^(3/2)= 6207

I got this wrong, but does anyone have any advice on how to solve this?
 

Attachments

  • physics problem.png
    physics problem.png
    2.8 KB · Views: 7,189
Last edited:
Physics news on Phys.org
littlejon said:

Homework Statement


Specify the strength of the electric field. Let r = 7.3cm .

Homework Equations


E1= (k*p)/(r^2+a^2)^3/2
Here k= constant used in Coulomb's law. a= half the length of dipole
p= dipole moment = charge x length of dipole. Charge = 3nC and length of dipole

The Attempt at a Solution


E1= (9E9)(3E-9)/(.146)^2+(.073)^2)^(3/2)= 6207

I got this wrong, but does anyone have any advice on how to solve this?
It appears that you used the amount of charge for p, rather than dipole moment.

## p=2aq ##
 
SammyS said:
It appears that you used the amount of charge for p, rather than dipole moment.

## p=2aq ##
Then ((9E9)(2*.073*3E-9))/((.146^2)+(.073)^2)^(3/2) = 906?
 
I'm surprised that a problem like this would have a be larger than r. Usually it's the other way around.

What is the complete problem you were given to solve? -- word for word.
 
SammyS said:
I'm surprised that a problem like this would have a be larger than r. Usually it's the other way around.

What is the complete problem you were given to solve? -- word for word.
What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?
Specify the strength of the electric field. Let r = 7.3cm .
Express your answer using two significant figures.
Thats word for word.
 
littlejon said:
What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?
Specify the strength of the electric field. Let r = 7.3cm .
Express your answer using two significant figures.
That's word for word.
In the figure, the charges have a separation, 2r . That is to say, a = r .

physics-problem-png.80347.png
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top