A particle with charge q1 = +4.4 µC is located atx = 0, and a second particle with charge q2 = −2.3 µC is located at x = 15 cm. Determine the location of a third particle with charge q3 = +3.6 µC such that the netelectric field at x = 25 cm is zero.
To determine the location of the third particle, we need to find the net electric field resulting from the first two particles and set it to zero at x = 25 cm.
First, let's calculate the magnitudes and directions of the electric fields created by each of the particles at x = 25 cm.
1. Electric field created by particle 1:
The electric field created by a point charge is given by the formula:
E₁ = k * |q₁| / r₁²,
Where E₁ is the electric field, k is the Coulomb constant (k = 9 × 10^9 N.m²/C²), |q₁| is the magnitude of charge q₁, and r₁ is the distance from charge q₁ to the location where we want to find the net electric field.
Given that q₁ = +4.4 µC and r₁ = 25 cm = 0.25 m:
E₁ = (9 × 10^9 N.m²/C²) * (4.4 × 10^-6 C) / (0.25 m)²
2. Electric field created by particle 2:
Following the same formula, the electric field created by particle 2 is:
E₂ = k * |q₂| / r₂²,
where q₂ = -2.3 µC and r₂ = 15 cm = 0.15 m:
E₂ = (9 × 10^9 N.m²/C²) * (2.3 × 10^-6 C) / (0.15 m)²
Next, let's add the electric fields together:
E_net = E₁ + E₂
Since we want the net electric field to be zero at x = 25 cm, the equation becomes:
E_net = 0
Plugging in the values, we get:
0 = E₁ + E₂
Solving this equation will give us the location of the third particle where the net electric field is zero.
To determine the location of the third particle, we need to find the distance from x = 25 cm where the net electric field is zero.
The net electric field at x = 25 cm will be zero if the electric fields due to the first and second particles at that point cancel each other out. 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 (k = 8.99 x 10^9 Nm^2/C^2), q is the charge of the particle, and r is the distance from the particle.
Let's calculate the electric fields due to the first and second particles at x = 25 cm:
Electric field due to q1 = k * (q1 / r1^2)
Electric field due to q2 = k * (q2 / r2^2)
Since we know the charges and distances from the particles, we can substitute those values into the equations:
Electric field due to q1 = (8.99 x 10^9 Nm^2/C^2) * (4.4 x 10^-6 C) / (25 cm)^2
Electric field due to q2 = (8.99 x 10^9 Nm^2/C^2) * (-2.3 x 10^-6 C) / (25 cm - 15 cm)^2
Now we can calculate the values of these electric fields:
Electric field due to q1 = 4.765 x 10^5 N/C
Electric field due to q2 = -1.311 x 10^6 N/C
Since the electric fields have opposite signs, they can cancel each other out and result in a net electric field of zero. For this to happen, the magnitudes of the electric fields need to be equal. Therefore,
|Electric field due to q1| = |Electric field due to q2|
4.765 x 10^5 N/C = 1.311 x 10^6 N/C
Now, we can solve for the distance r3, which is the distance from the third particle to x = 25 cm. Rearranging the equation for the electric field, we get:
r3^2 = (k * |q|) / |Electric field|
Substituting the values, we have:
r3^2 = (8.99 x 10^9 Nm^2/C^2 * (3.6 x 10^-6 C)) / (4.765 x 10^5 N/C)
Now we can solve for r3:
r3^2 = 6.811 x 10^-3 m^2
Taking the square root of both sides, we get:
r3 = 0.0826 m
Therefore, the location of the third particle with charge q3 = +3.6 µC such that the net electric field at x = 25 cm is zero is at a distance of approximately 0.0826 meters (or 82.6 cm) from x = 25 cm.