Answer:
E = 0.437 N/C
Explanation:
Given that,
Charge,
Electric force,
Let the strength of the electric field is E. We know that, the electric force is given by :
F = qE
Where
E is the electric field strength
So, the strength of the electric field is equal to 0.437 N/C.
Answer:
a) At 0.20 m, the magnitude of the field is 675.0 kV
The direction of the field is acting outwards from the center of the charged spheres
b) At 0.10 m, the magnitude of the field is 135 kV
The direction is acting outwards from the center of the charged spheres
c) At 0.025 m
The magnitude of the field, V = -270.0 kV
The direction of the field is inwards, towards the center of the charged spheres
Explanation:
The charged spherical shell parameters are;
The charge on the inner sphere, q₁ = -1.50 × 10⁻⁶ C
The radius of the inner shell, R₁ = 0.050 m
The charge on the outer sphere, q₂ = +4.50 × 10⁻⁶ C
The radius of the outer shell, R₂ = 0.15 m
Let 'r', represent the distance at which the electric field is measured, the following relationships can be obtained;
When r < R₁ < R₂,
When R₁ < r < R₂,
When R₁ < R₂ < r,
a) When r = 0.20 m, we have;
R₁ < R₂ < r, therefore
By plugging in the values, we get;
Therefore, the magnitude of the field, V = 675.0 kV
The direction of the field is outwards
b) When r = 0.10 m, we have;
When R₁ < r < R₂, therefore;
By plugging in the values, we get;
Therefore, the magnitude of the field, V = 135 kV
The direction of the field is outwards from the center
c) When r = 0.025 m, we have;
When r < R₁ < R₂, therefore;
By plugging in the values, we get;
Therefore, the magnitude of the field, V = -270.0 kV
The direction of the field is inwards, towards the center of the charged spheres.