Finding the electric field vector along the bisector

[dk321]

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?

 

[tiny-tim]

welcome to pf!

hi dk321! welcome to pf!

(try using the X² and B buttons just above the Reply box )

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?

 

[dk321]

why is it j ? the two charges line in horizontal direction so along the perpendicular bisector would be in vertical direction right?

 

[tiny-tim]

why is it j ? the two charges line in horizontal direction so along the perpendicular bisector would be in vertical direction right?

yes, j

 

[darkxponent]

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!

 

[darkxponent]

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?