Final answer:
The magnitude of the electric field at a point 2.5 cm from the symmetry axis of the two surfaces is 151.2 N/C.
Step-by-step explanation:
To determine the magnitude of the electric field at a point 2.5 cm from the symmetry axis of the two surfaces, we can use the principle of superposition. The electric field due to the cylindrical surface can be calculated using the formula:
E1 = (k * λ1) / r1
where
k is Coulomb's constant (9.0 x 10^9 Nm²/C²),
λ1 is the charge density of the cylindrical surface (-30 pC/m²),
and r1 is the distance from the point to the cylindrical surface (2.5 cm).
Similarly, the electric field due to the coaxial surface can be calculated using:
E2 = (k * λ2) / r2
where
λ2 is the charge density of the coaxial surface (-12 pC/m²),
and r2 is the distance from the point to the coaxial surface (2.5 cm).
By adding the magnitudes of E1 and E2, we can obtain the total electric field at the point.
Let's calculate:
E1 = (9.0 x 10^9 Nm²/C² * (-30 x 10^-12 C/m²)) / (2.5 x 10^-2 m) = -108 N/C
E2 = (9.0 x 10^9 Nm²/C² * (-12 x 10^-12 C/m²)) / (2.5 x 10^-2 m) = -43.2 N/C
The total electric field is the sum of the magnitudes, which gives:
E_total = |-108 N/C| + |-43.2 N/C| = 151.2 N/C
Therefore, the magnitude of the electric field at a point 2.5 cm from the symmetry axis of the two surfaces is 151.2 N/C.