Two point charges, 4.0 µC and -2.0 µC, are placed 4.4 cm apart on the x axis, such that the -2.0 µC charge is at x = 0 and the 4.0 µC charge is at x = 4.4 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
cm
0 < x < 4.4 cm
cm
4.4 cm < x
cm
(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
cm
0 < x < 4.4 cm
cm
4.4 cm < x
cm
A charge of 2 µC and 4 µC are 4 cm apart. What is the force between them?
To find the points along the x-axis where the electric field is zero, we can use the principle that the electric field due to the two charges will cancel out when the magnitudes of the electric field due to each charge are equal.
Let's consider a point P on the x-axis, such that the distance from the -2.0 µC charge to point P is x, and the distance from the 4.0 µC charge to point P is 4.4 cm - x.
According to Coulomb's Law, the electric field due to a point charge is given by the equation:
E = k * q / r^2
Where E is the electric field, k is the electrostatic constant (9 * 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge to the point where the electric field is being measured.
For the -2.0 µC charge at x = 0, the electric field at point P is:
E1 = k * (-2.0 µC) / x^2
Similarly, for the 4.0 µC charge at x = 4.4 cm, the electric field at point P is:
E2 = k * (4.0 µC) / (4.4 cm - x)^2
To find the points where the electric field is zero, we need to solve the equation:
E1 = E2
k * (-2.0 µC) / x^2 = k * (4.0 µC) / (4.4 cm - x)^2
Now, we can cancel out the electrostatic constant (k) and simplify the equation:
(-2.0 µC) / x^2 = (4.0 µC) / (4.4 cm - x)^2
Now, we can cross multiply and rearrange the equation:
(-2.0 µC) * (4.4 cm - x)^2 = (4.0 µC) * x^2
Simplifying further:
(-2.0 µC) * (19.36 cm^2 - 8.8 cm * x + x^2) = (4.0 µC) * x^2
Expanding and collecting like terms:
(-38.72 µC * cm^2) + (17.6 µC * cm * x) - (2.0 µC * x^2) = 4.0 µC * x^2
Bringing all terms to one side of the equation:
6.0 µC * x^2 + 17.6 µC * cm * x - 38.72 µC * cm^2 = 0
Now, we can solve this quadratic equation for x using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Where a = 6.0 µC, b = 17.6 µC * cm, and c = -38.72 µC * cm^2.
Plugging the values into the formula and solving for x will give us the points along the x-axis where the electric field is zero.
To find the points along the x-axis where the electric potential is zero, we need to use the equation for electric potential due to a point charge, which is:
V = k * q / r
Similar to the previous approach, we can calculate the electric potential due to the -2.0 µC charge at point P and the electric potential due to the 4.0 µC charge at point P and equate them to zero. Then, solving for x will give us the points where the potential is zero.
However, it's worth noting that in both cases (electric field and electric potential), the equations become quite complex due to the presence of squares and fractional units. Mathematica software or other numerical methods may be more suitable for solving these equations.