Download Finding the net electric field on the perpendicular bisector of the line

Finding the net electric field on the perpendicular bisector of the line joining two charges
Refer to the diagram below. Charges q1 and q2 are separated by distance d. The perpendicular bisector of
the line joining the charges is shown. A point P is indicated on this line. The problem is to determine the
magnitude and direction of the net electric field at P due to the two charges.
Solution
Notes
+y
q1
d/2
P
(x, 0)
0
d/2
+x
The coordinate system is set up to exploit the
symmetry of the situation. The line joining the
charges is the y-axis, and the origin is the
midpoint. The x-axis is the perpendicular bisector
of the line joining the charges. The charges are
positioned at (0, d/2) and (0, -d/2) respectively.
Point P is positioned at (x,0).
q2
-y
The following is also given:
+y
q1
q1 = +q
EP2
r
d/2
θ
θ
0
d/2
P
(x, 0)
r
q2
-y
Enet , Px = 2 EP 2 cos θ
Enet , Py = 0
Enet , Px = 2 EP 2 cos θ
2k q2 x
=
r2 r
2kqx
= 3
r
2kqx
=
3/ 2
⎡⎣ x 2 + (d / 2) 2 ⎤⎦
θ
θ
+x
EP1
q2 = +q
where q represents a quantity of positive charge.
This enables us to decide on the directions for the
E-field vectors. The two vectors extend from the
point P where the net field is to be found since the
electric force on a positive test charge at P would
be away from either charge.
The distance from either charge to point P is
represented by the symbol r.
The symmetry of the situation and the way we set
up the coordinate system makes it simple to
calculate the components of the net field. The xcomponents of EP1 and EP2 are equal so the net Efield is twice either one. The y-components are
equal and opposite, so they add to 0.
Next we substitute the expression for the field of a
point charge.
From trigonometry we have cos θ = x / r , and
from the Pythagorean theorem, we have
r = x2 + (½d )2 . Substituting, we obtain the final
expression.