Three concentric rings and one point charge are in the same plane as shown in the above diagram. The linear charge densities of the rings are ë1 = 2.60nC/m, ë2 = -1.10nC/m, and ë3 = 2.20nC/m with corresponding radii of r1 = 86.0cm, r2 = 64.0cm, and r3 = 46.0cm. The point charge at the center of the rings has a net charge of q = -3.10nC. What is the magnitude of the electric field at a point along the axis that runs through the center of the rings, which is a distance a = 2.40m away from the center of the rings?
To find the magnitude of the electric field at a point along the axis passing through the center of the rings, we can use the principle of superposition. The electric field due to each individual charge is calculated separately, and then their contributions are added together.
The electric field due to a point charge is given by Coulomb's law:
E = k * (q / r^2)
where E is the electric field, k is the Coulomb's constant (9.0 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge.
Let's calculate the electric field due to each component individually:
1. Electric Field due to the point charge at the center:
The charge at the center has a charge of q = -3.10nC. Since the distance (a) is given as 2.40m, the electric field due to this charge can be calculated as:
E1 = (9.0 x 10^9 N m^2/C^2) * (-3.10 x 10^-9 C) / (2.40 m)^2
2. Electric Field due to the first ring:
The linear charge density of the first ring is given as ë1 = 2.60 nC/m. We need to convert it to total charge by multiplying it with the circumference of the ring, given by 2πr1, where r1 is the radius of the first ring (86.0 cm). The electric field is then given by:
E2 = (9.0 x 10^9 N m^2/C^2) * (2.60 x 10^-9 C/m) * (2π * 0.86 m)
3. Electric Field due to the second ring:
Similarly, we need to convert the linear charge density ë2 = -1.10 nC/m to total charge by multiplying it with the circumference of the ring, given by 2πr2, where r2 is the radius of the second ring (64.0 cm). The electric field is then given by:
E3 = (9.0 x 10^9 N m^2/C^2) * (-1.10 x 10^-9 C/m) * (2π * 0.64 m)
4. Electric Field due to the third ring:
We need to convert the linear charge density ë3 = 2.20 nC/m to total charge by multiplying it with the circumference of the ring, given by 2πr3, where r3 is the radius of the third ring (46.0 cm). The electric field is then given by:
E4 = (9.0 x 10^9 N m^2/C^2) * (2.20 x 10^-9 C/m) * (2π * 0.46 m)
Finally, to find the total electric field at the given distance, a = 2.40 m, we sum up all the individual contributions:
E_total = E1 + E2 + E3 + E4
The magnitude of the electric field is then given by the absolute value of E_total.