Finding the electric field vector along the bisector

AI Thread Summary
The discussion focuses on finding the electric field vector along the perpendicular bisector of two equal positive charges separated by distance D. The initial equation presented for the electric field was corrected to indicate that the vector direction should be along the vertical (j direction) rather than horizontal (i direction). Participants confirmed that the first part of the question was addressed correctly, while the second part requires using calculus to determine where the electric field is maximized as a function of distance. The importance of symmetry and the principle of superposition in solving the problem was emphasized. The conversation highlights the need for clarity in vector direction and the application of mathematical techniques to find maxima.
dk321
Messages
4
Reaction score
0
Two equal positive charges are held fixed and separated by distance D. Find the electric field
vector along their perpendicular bisector. Then find the position relative to their center where
the field is a maximum.

so i started with symmetry and principle of superposition

and got to this answer
E = (1/(4πε))*q*D i^/( y^2 + (D^2)/4)^(3/2)

y stand for the distance from the field to the origin

is this equation the right answer for the first question? Also for the second question what changes when E is a maximum?
 
Physics news on Phys.org
welcome to pf!

hi dk321! welcome to pf! :smile:

(try using the X2 and B buttons just above the Reply box :wink:)
dk321 said:
E = (1/(4πε))*q*D i/( y2 + (D2)/4)3/2

nooo, draw a diagram … it isn't along i, i'ts along j, isn't it? :wink:
 
why is it j ? the two charges line in horizontal direction so along the perpendicular bisector would be in vertical direction right?
 
dk321 said:
why is it j ? the two charges line in horizontal direction so along the perpendicular bisector would be in vertical direction right?

yes, j :confused:
 
dk321 said:
Two equal positive charges are held fixed and separated by distance D. Find the electric field
vector along their perpendicular bisector. Then find the position relative to their center where
the field is a maximum.

so i started with symmetry and principle of superposition

and got to this answer
E = (1/(4πε))*q*D i^/( y^2 + (D^2)/4)^(3/2)

y stand for the distance from the field to the origin

is this equation the right answer for the first question? Also for the second question what changes when E is a maximum?

The first part is right.

Hint for second part: E is a function of 'd'. And you have to find the maxima of E(d). How about using some calculus!
 
dk321 said:
why is it j ? the two charges line in horizontal direction so along the perpendicular bisector would be in vertical direction right?

And the vertical direction is 'j'. Isnt it?
 

Thread 'Errors and significant figures'

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...
View full post »
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanging mass'

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...
View full post »

Similar threads

Electric field created by point charges
Replies
6
Views
752
Find the electric field between 2 finite discs
Replies
16
Views
1K
How to find vectors and bisectors of an electric field
Replies
9
Views
1K
Two equal positive charges are held fixed and separated by d
Replies
9
Views
3K
Direction and magnitude of electric field produced by current change
Replies
8
Views
981
Finding the position where the electric field is zero
Replies
2
Views
1K
Why does electric potential energy increase if you move against the field?
Replies
6
Views
1K
Poynting vector and electric field
Replies
4
Views
2K
Why is electric field at the center of a charged disk not zero?
Replies
4
Views
3K
Electric field of point along the central vertical axis of a triangle
Replies
3
Views
801

Hot Threads

Recent Insights

Back
Top