Download Example 16-6 Where Is the Electric Field Zero?

Example 16-6 Where Is the Electric Field Zero?
A point charge Q1 = +4.00 nC (1 nC = 1 nanocoulomb = 1029 C) is placed 0.500 m to the left of a point charge
Q2 = +9.00 nC. Find the position between the two point charges where the net electric field is zero.
Set Up
At any point between the two charges, the electric
s1 due to Q1 points to the right (away from
field E
s2 due
this positive charge) and the electric field E
to Q2 points to the left (away from this positive
charge). We want to find the point P where the total
s1 + E
s2 equals zero. We’ll use the symbol
s = E
field E
D for the 0.500-m distance between the two charges
and x for the distance from Q1 to point P: Then the
distance from Q2 to point P is D 2 x. Our goal is to
s = 0.
find the value of x for which E
Solve
s1 (to
If the net electric field at P is zero, then E
s2 (to the left) must have equal magthe right) and E
nitudes so that these two vectors cancel. Use Equation 16-4 to write this statement in equation form.
Magnitude of the electric field Q1 = +4.00 nC E2 P E1
due to a point charge Q:
E =
Q2 = +9.00 nC
x
k Q 
r2
D = 0.500 m
(16-4)
Total electric field:
s = E
s1 + E
s 2
E
(16-5)
In order for the net electric field at P to be zero,
E1 = E2
The distance from Q1 to P is x, and the distance from Q2 to P is
D 2 x, so from Equation 16-4
E1 =
k Q1 
x
2
and E 2 =
k Q2 
1D - x2 2
If these are equal to each other,
k Q1 
x
Solve this equation for x.
2
=
k Q2 
1D - x2 2
The factors of k cancel in the above equation, so
 Q1 
 Q2 
=
or
x2
1D - x2 2
 Q1  1D - x2 2 =  Q2  x 2
Multiply out the quantity (D 2 x)2 and rearrange:
 Q1  1D 2 - 2Dx + x 2 2 =  Q2  x 2
1  Q2  -  Q1  2x 2 + 12D Q1  2x + 1 -D 2  Q1  2 = 0
We can simplify this if we divide through by  Q1  :
a
 Q2 
 Q1 
- 1bx 2 + 2Dx + 1 -D 2 2 = 0
This is a quadratic equation of the form ax2 + bx + c = 0, with
a =  Q2  >  Q1  - 1 =  9.00 nC  >  4.00 nC  - 1 = 1.25,
b = 2D = 2(0.500 m) = 1.00 m, and c = 2D2 = 2(0.500 m)2 =
20.250 m2. The solutions are
x =
=
=
-b { 2b 2 - 4ac
2a
- 11.00 m2 { 211.00 m2 2 - 411.252 1 -0.250 m2 2
211.252
- 11.00 m2 { 22.25 m2
2.50
= +0.200 m or 21.00 m
=
- 11.00 m2 { 11.50 m2
2.50
We want a positive value of x to correspond to a point to the right of
Q1, so the solution we want is x = +0.200 m. We conclude that point
P is a distance x = 0.200 m to the right of charge Q1 and a distance
D 2 x = 0.500 m 2 0.200 m = 0.300 m to the left of charge Q2.
Reflect
Because charge Q1 is smaller than Q2, the location of the point where
E1
Q1 = +4.00 nC
Q2 = +9.00 nC
the net electric field is zero must be closer to Q1 than to Q2. That’s just
what we found. You can check the result x = 0.200 m by substituting
1.00 m
D = 0.500 m
E2
this value into the above expressions for E1 and E2 and confirming that
E1 = E2 for this value of x.
But what’s the significance of the second solution, x = 21.00 m? This refers to a point 1.00 m to the left of charge
Q1 and 1.00 m + 0.500 m = 1.50 m to the left of charge Q2. Our calculation shows that this point E1 is equal to E2.
s2 both point to the left (both fields point away from the positive charges that produce
s1 and E
However, at this point E
s2 do not cancel, and the total field is not zero.
s1 and E
them). So at this point the electric fields E