Two point charges, 5.0 µC and -2.0 µC, are placed 5.0 cm apart on the x axis, such that the -2.0 µC charge is at x = 0 and the 5.0 µC charge is at x = 5.0 cm. At what point(s) along the x axis is the electric field zero?
I'm not exactly sure how i should go by solving this.
Well, it seems like the electric field is having a bit of an identity crisis. It can't decide whether it wants to be positive or negative! But don't worry, I'm here to help you solve this conundrum.
The electric field at any point in space is the vector sum of the electric fields due to each individual charge. In order for the total electric field to be zero, the magnitudes of the electric fields due to each charge must be equal.
Let's start by calculating the electric field due to the 5.0 µC charge at a point along the x-axis. We can use Coulomb's law for this:
E1 = k * (Q1/r^2)
where E1 is the electric field due to the 5.0 µC charge, k is the electrostatic constant, Q1 is the charge, and r is the distance from the charge.
Using the given values, we have:
E1 = (9 x 10^9 N*m^2/C^2) * (5.0 x 10^-6 C) / (0.05 m)^2
E1 ≈ 900 N/C
Now, let's calculate the electric field due to the -2.0 µC charge at a point along the x-axis. Again, we can use Coulomb's law:
E2 = k * (Q2/r^2)
Using the given values, we have:
E2 = (9 x 10^9 N*m^2/C^2) * (-2.0 x 10^-6 C) / (0.05 m)^2
E2 ≈ -720 N/C
Since the electric fields are vectors, in order for them to cancel each other out and result in a net electric field of zero, their magnitudes must be equal. However, in this case, E1 = 900 N/C and E2 = -720 N/C. So, at no point along the x-axis will the electric field be zero.
Looks like these charges are quite the odd couple, never quite able to see eye to eye on things, even in terms of electric fields.
To find the point(s) along the x-axis where the electric field is zero, we need to consider the electric fields created by each of the point charges and their superposition.
The electric field created by a point charge q at a point P is given by Coulomb's law:
E = k * q / r^2
where E is the electric field, k is the electrostatic constant (k = 9 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge to the point P.
In this case, we have two point charges, one with a positive charge (+5.0 µC) and the other with a negative charge (-2.0 µC). Let's call the positive charge Q1 and the negative charge Q2.
At a point P along the x-axis, the electric field created by Q1 (E1) and Q2 (E2) at that point will be in opposite directions due to their opposite charges. Thus, the total electric field at point P will be the vector sum of E1 and E2:
E_total = E1 + E2
To find the point(s) where the electric field is zero, we set E_total = 0:
E1 + E2 = 0
Let's substitute the expressions for the electric fields:
(k * Q1 / r1^2) + (k * Q2 / r2^2) = 0
Simplifying the equation:
Q1 / r1^2 + Q2 / r2^2 = 0
For the given problem, Q1 = 5.0 µC and Q2 = -2.0 µC. The distances r1 and r2 are the distances from P to each of the charges.
Let's denote the distance from P to Q1 as x, and the distance from P to Q2 as (5 - x) cm, since Q1 is at x = 5.0 cm.
Now we can substitute the values into the equation:
(5.0 µC / x^2) + (-2.0 µC / (5 - x)^2) = 0
Solving this equation will give us the value(s) of x where the electric field is zero.
To find the point(s) along the x-axis where the electric field is zero, we need to consider the concept of superposition of electric fields.
The electric field created by a point charge can be calculated using Coulomb's Law:
E = k * (q / r^2)
where E is the electric field, k is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), q is the charge, and r is the distance from the charge.
In this case, we have two point charges: +5.0 µC and -2.0 µC. The electric field created by each charge will have a direction along the x-axis. Since the electric field is a vector quantity, it can be either positive or negative depending on the direction.
To find the point(s) where the electric field is zero, we need to consider the superposition of electric fields from both charges. The net electric field at a particular point is the vector sum of the electric fields created by each individual charge.
At a point along the x-axis, the electric field created by the +5.0 µC charge will be directed away from the charge, while the electric field created by the -2.0 µC charge will be directed towards the charge. The magnitude of the electric field is given by Coulomb's Law.
To find the point(s) where the net electric field is zero, we need the magnitudes of the electric fields created by both charges to be equal. This implies that the distances from this point to the two charges are in inverse proportion to the magnitudes of the charges.
Let's solve for the position where the electric field is zero:
E1 = k * (5.0 µC) / (x^2) (electric field at x)
E2 = k * (-2.0 µC) / ((5.0 cm - x)^2) (electric field at 5 cm - x)
Since the electric field magnitudes must be equal, we can set up the equation:
k * (5.0 µC) / (x^2) = k * (-2.0 µC) / ((5.0 cm - x)^2)
Simplifying and canceling out the electrostatic constant k:
(5.0 µC) / (x^2) = (-2.0 µC) / ((5.0 cm - x)^2)
Cross-multiplying:
5.0 µC * ((5.0 cm - x)^2) = -2.0 µC * (x^2)
Expanding and rearranging:
25.0 cm^2 - 10.0 cm * x + x^2 = -2.0 cm^2
Adding 2.0 cm^2 to both sides:
x^2 - 10.0 cm * x + 27.0 cm^2 = 0
This is a quadratic equation, we can solve it by factoring, completing the square, or using the quadratic formula. Factoring might not be straightforward in this case, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation:
a = 1, b = -10.0 cm, c = 27.0 cm^2
Substituting into the quadratic formula:
x = (10.0 cm ± √((-10.0 cm)^2 - 4(1)(27.0 cm^2))) / (2(1))
Simplifying:
x = (10.0 cm ± √(100.0 cm^2 - 108.0 cm^2)) / 2
x = (10.0 cm ± √(-8.0 cm^2)) / 2
Here, we encounter an issue. The square root of a negative number is not meaningful in the context of this problem. Therefore, there are no real solutions for x, which means there are no points along the x-axis where the electric field is precisely zero.
In summary, there are no points along the x-axis where the electric field is zero when two point charges, 5.0 µC and -2.0 µC, are placed 5.0 cm apart.
First of all, recognize that the point cannot be between x = 0 and x = 5, because the E fields due to the two particles will be in the same direction and cannot cancel each other. In the x > 5 region, the E field due to the -2 uC charge will be less everywhere, so camncellation is not possible. The location will have to be in the x <0 region.
In the x < 0 region,
-2 k/ x^2 + 5k /(5-x)^2 = 0
5/(5-x)^2 = 2/x^2
5x^2 = 50 -20x +2x^2
3x^2 +20x -45 = 0
Take the root x that is < 0. This does not factor easily.