A very thin hollow conducting sphere has a radius of R=0.12 m. The center of the sphere is at S. A charge of q=7×10−6 C is present at point P which is located d=0.09 m from S. Assume that this charge stays at this position (thus it cannot move). We draw a line from S to P and extend it beyond the radius of the sphere.
What is the Electric field strength (V/m) at point D which is located on the line from SP at a distance of ℓ=0.38 m from S (thus 0.29 m from P)? Note that the graph is not drawn to scale
To find the electric field strength (E) at point D, we can use the principle of superposition. This principle states that the total electric field at a point due to multiple charges is the vector sum of the electric fields produced by each individual charge.
First, let's calculate the electric field at point D due to the charge at point P. The electric field (E1) at D due to a single point charge can be calculated using Coulomb's law:
E1 = k * (|q| / r^2)
Where:
- k is the Coulomb's constant, approximately 9 × 10^9 Nm^2/C^2
- |q| is the magnitude of the charge at P, given as 7 × 10^-6 C
- r is the distance from P to D, given as 0.29 m
Substituting the values into the equation:
E1 = (9 × 10^9 Nm^2/C^2) * (7 × 10^-6 C) / (0.29 m)^2
Calculating this, E1 ≈ 7.42 × 10^5 N/C
Now, let's calculate the electric field at point D due to the hollow conducting sphere.
For any point outside a hollow conductor, the electric field inside the conductor is zero. This is because charges inside the conductor redistribute themselves due to the presence of an external charge, cancelling out the electric field within the conductor.
Therefore, the electric field (E2) at point D due to the hollow conducting sphere is zero.
Finally, the total electric field (E) at point D is the vector sum of E1 and E2:
E = E1 + E2 = 7.42 × 10^5 N/C + 0 N/C
Hence, the electric field strength at point D is approximately 7.42 × 10^5 N/C.