Two spherical shells have a common center. A -3.00 × 10-6 C charge is spread uniformly over the inner shell, which has a radius of 0.050 m. A +4.00 × 10-6 C charge is spread uniformly over the outer shell, which has a radius of 0.15 m. Find the magnitude and direction of the electric field at a distance (measured from the common center) of
To find the magnitude and direction of the electric field at a distance from the common center, we can use the principle of superposition. The electric field at that point is the vector sum of the electric fields produced by the two shells.
To calculate the electric field produced by each shell, we can use the equation:
E = k(q / r^2)
where E is the electric field, k is Coulomb's constant (9 × 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge.
For the inner shell with charge -3.00 × 10^-6 C, we can calculate the electric field at the given distance by substituting the values into the equation:
E_inner = (9 × 10^9 N m^2/C^2) * (-3.00 × 10^-6 C) / (0.050 m)^2
E_inner = -108 N/C (note: the negative sign represents the direction of the electric field)
For the outer shell with charge +4.00 × 10^-6 C, we can calculate the electric field at the given distance by substituting the values into the equation:
E_outer = (9 × 10^9 N m^2/C^2) * (4.00 × 10^-6 C) / (0.15 m)^2
E_outer = 32 N/C (since the outer shell has a positive charge, the direction of the electric field is away from it)
Finally, we can find the total electric field by adding the electric fields produced by each shell:
E_total = E_inner + E_outer
E_total = -108 N/C + 32 N/C
E_total = -76 N/C
So, the magnitude of the electric field at the given distance is 76 N/C, and the direction is towards the inner shell (opposite to the direction of the electric field produced by the outer shell).