Two point charges are placed along a horizontal axis with the following values and positions: +6.0 µC at x = 0 cm and -24.0 µC at x = 60 cm. At what point along the x axis is the electric field zero?
It won't be between the charges, because both charges produce an E-field in the same direction there. Look for it to be along the -X axis, at a point x = -a, where
6/a^2 = 24/(60 + a)^2
(60+a)/a = sqrt(4) = 2
60/a + 1 = 2
a = 60 cm
Well, it seems like those point charges are really charged up about their positions! So, let's see if we can bring some balance into the electric field.
The electric field at a point due to a point charge is given by the equation:
E = k * Q / r^2
Where E is the electric field, k is the Coulomb's constant (8.99 x 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge to the point. In this case, we have two charges, so we need to find the point where the electric fields balance out.
For the electric fields to cancel each other out, the magnitudes of the two electric fields must be equal:
E1 = E2
k * Q1 / r1^2 = k * Q2 / r2^2
Using the given values: 6.0 µC and -24.0 µC, and the corresponding distances: 0 cm and 60 cm, we can rearrange the equation and solve for r:
Q1 / r1^2 = -Q2 / r2^2
Q1 * r2^2 = -Q2 * r1^2
r2^2 = (Q2 * r1^2) / Q1
r2 = sqrt((Q2 * r1^2) / Q1)
Plugging in the values, we get:
r2 = sqrt((-24.0 µC * (0.60 m)^2) / 6.0 µC)
Calculating that out, we find:
r2 ≈ sqrt(144) m
So the electric field will be zero at the point approximately 12 m clownwards from the -24.0 µC charge.
To find the point along the x-axis where the electric field is zero, we can use the principle of superposition.
The electric field at a point due to multiple charges is the vector sum of the electric fields produced by each charge at that point.
Let's label the position along the x-axis where the electric field is zero as x_zero.
Now, let's calculate the electric field at x_zero due to the +6.0 µC and -24.0 µC charges individually.
Electric field due to the +6.0 µC charge:
E1 = k * |q1| / (x_zero - x1)^2
Electric field due to the -24.0 µC charge:
E2 = k * |q2| / (x_zero - x2)^2
In the above equations:
k is the electrostatic constant (k = 8.99 × 10^9 N m^2/C^2)
q1 = +6.0 µC = 6.0 × 10^-6 C
q2 = -24.0 µC = -24.0 × 10^-6 C
x1 = 0 cm = 0 m
x2 = 60 cm = 0.6 m
Now, we need to find the value of x_zero for which E1 + E2 = 0.
k * |q1| / (x_zero - x1)^2 + k * |q2| / (x_zero - x2)^2 = 0
Substituting the values we have:
(8.99 × 10^9 N m^2/C^2) * (6.0 × 10^-6 C) / (x_zero - 0 m)^2 + (8.99 × 10^9 N m^2/C^2) * (-24.0 × 10^-6 C) / (x_zero - 0.6 m)^2 = 0
Simplifying the equation and solving for x_zero will give us the position where the electric field is zero.
To find the point along the x-axis where the electric field is zero, we can make use of the principle of superposition. According to this principle, the total electric field at any point is the vector sum of the individual electric fields created by each point charge.
Let's denote the distance from the first charge (+6.0 µC) to the unknown point as x, and the distance from the second charge (-24.0 µC) to the same point as (60 - x) cm.
The electric field created by a point charge in free space is given by Coulomb's law:
E = k * q / r^2,
where E is the electric field, k is the electrostatic constant (8.99 × 10^9 N·m^2/C^2), q is the charge of the point charge, and r is the distance from the point charge to the point where the field is measured.
Using this formula, we can calculate the magnitudes of the electric fields created by each point charge at the unknown point.
For the charge of +6.0 µC at x = 0 cm, the distance from the charge to the unknown point is x cm, so the electric field created by it is:
E1 = (8.99 × 10^9 N·m^2/C^2) * (6.0 × 10^(-6) C) / (x × 10^(-2) m)^2.
For the charge of -24.0 µC at x = 60 cm, the distance from the charge to the unknown point is (60 - x) cm, so the electric field created by it is:
E2 = (8.99 × 10^9 N·m^2/C^2) * (-24.0 × 10^(-6) C) / ((60 - x) × 10^(-2) m)^2.
Now, since we want the total electric field to be zero at the unknown point, the magnitudes of the electric fields created by each charge must be equal:
|E1| = |E2|.
Therefore, we can set up the equation:
(8.99 × 10^9 N·m^2/C^2) * (6.0 × 10^(-6) C) / (x × 10^(-2) m)^2 = (8.99 × 10^9 N·m^2/C^2) * (-24.0 × 10^(-6) C) / ((60 - x) × 10^(-2) m)^2.
Simplifying the equation, we get:
(6.0 × 10^(-6)) / x^2 = (-24.0 × 10^(-6)) / (60 - x)^2.
Cross-multiplying and rearranging terms, we have:
x^2(60 - x)^2 = 4(60 - x)^2.
Dividing both sides by (60 - x)^2, we obtain:
x^2 = 4.
Taking the square root of both sides, we get two solutions:
x = ± 2.
Since we are considering the positive x-axis, the point along the x-axis where the electric field is zero is x = 2 cm.
Therefore, the electric field is zero at x = 2 cm.