Two point charges, 6.0 µC and -2.0 µC, are placed 6.0 cm apart on the x axis, such that the -2.0 µC charge is at x = 0 and the 6.0 µC charge is at x = 6.0 cm.
(a) At what point(s) along the x axis is the electric field zero? (If there is no point where E = 0 in a region, enter "0" in that box.)
x < 0 ___________
0 < x < 6.0 cm____________
6.0 cm < x____________
(b) At what point(s) along the x axis is the potential zero? Let V = 0 at r = . (If there is no point where V = 0 in a region, enter "0" in that box.)
x < 0 ____________
0 < x < 6.0 cm______________
6.0 cm < x________________
I don't get it..
could u please explain it
You couldn't be less helpful
To find the points along the x-axis where the electric field (E) is zero and where the potential (V) is zero, we need to calculate the electric field and potential at different points.
First, let's define some variables:
- q1 = 6.0 µC (charge at x = 6.0 cm)
- q2 = -2.0 µC (charge at x = 0)
- d = 6.0 cm (separation between the charges)
- x = position along the x-axis
(a) Electric Field (E):
The electric field at a point is given by the formula:
E = (k * |q1|) / (r1^2) - (k * |q2|) / (r2^2)
where:
- k = Coulomb's constant (8.99 x 10^9 N m^2 / C^2)
- r1 = distance from point to q1
- r2 = distance from point to q2
We want to find the points where E = 0.
Set E = 0 and solve for x:
0 = (k * |q1|) / (x^2) - (k * |q2|) / ((d-x)^2)
Simplifying the equation will give the point(s) at which the electric field is zero.
(b) Potential (V):
The potential at a point is given by the formula:
V = (k * q1) / r1 - (k * q2) / r2
Similarly, set V = 0 and solve for x:
0 = (k * q1) / x - (k * q2) / (d-x)
Solving this equation will provide the point(s) at which the potential is zero.
By calculating the values and solving the equations, you can find the answers for both (a) and (b).
so figure the E from each charge, and add them, set to zero, and solve.
It is obvious that between the charges, E will point to the left, and there is no solution there.
I will set up the problem for the x<0
E=0=-2k/x^2 +6k/(absx+6)^2
2k/x^2=6k/(x+6)^2
take the sqrt of each side.
x=(abs x+6)/sqrt 3
absx *sqrt3-absx=6
solve for x, remembering x is the distance from zero to the left.
so figure the E from each charge, and add them, set to zero, and solve.
I am not going to do all that algebra for you. I will be happy to critique or check your work.