Question

A conducting solid sphere of radius \(20.0 \mathrm{~cm}\) is located with its center at the origin of a three-dimensional coordinate system. A charge of \(0.271 \mathrm{nC}\) is placed on the sphere. a) What is the magnitude of the electric field at point \((x, y, z)=\) \((23.1 \mathrm{~cm}, 1.1 \mathrm{~cm}, 0 \mathrm{~cm}) ?\) b) What is the angle of this electric field with the \(x\) -axis at this point? c) What is the magnitude of the electric field at point \((x, y, z)=\) \((4.1 \mathrm{~cm}, 1.1 \mathrm{~cm}, 0 \mathrm{~cm}) ?\)

Short answer

Question: For a uniformly charged solid sphere of radius 20 cm with a charge of \(0.271 \times 10^{-9} \mathrm{C}\), calculate: a) The electric field at point A (23.1 cm, 1.1 cm, 0 cm); b) The angle between the electric field at point A and the x-axis; c) The electric field at point C (4.1 cm, 1.1 cm, 0 cm).

Step-by-step solution

For Point A (23.1 cm, 1.1 cm, 0 cm)

Calculate the distance from the origin (center of the sphere): $$r_A = \sqrt{(23.1)^2 + (1.1)^2} \approx 23.21 \mathrm{~cm} > 20 \mathrm{~cm}$$The point is outside the sphere.

For Point C (4.1 cm, 1.1 cm, 0 cm)

Calculate the distance from the origin (center of the sphere): $$r_C = \sqrt{(4.1)^2 + (1.1)^2} \approx 4.26 \mathrm{~cm} < 20 \mathrm{~cm}$$The point is inside the sphere. Step 2: Calculate the electric field

For Point A (Electric Field Magnitude)

Given charge Q = \(0.271 \times 10^{-9} \mathrm{C}\). To calculate the electric field E at point A, we use the formula: $$E_A = \frac{kQ}{r_A^2} = \frac{8.99\times10^9 \times 0.271 \times 10^{-9}}{(23.21\times10^{-2})^2} \approx 4.84 \times 10^3 \mathrm{N/C}$$

For Point B (Electric Field Angle)

The angle between the electric field and x-axis can be found using the x-component of the electric field vector: $$cos \theta = \frac{E_{x}}{E_A}$$ The x-component of the electric field is proportional to the x-component of the distance from the origin: $$E_{x} = E_A \frac{x_{A}}{r_A} = 4.84 \times 10^3 \frac{23.1 \mathrm{~cm}}{23.21 \mathrm{~cm}}$$ Now, we can find the angle $$\theta = arccos{\left(\frac{E_{x}}{E_A}\right)} = arccos{\left(\frac{23.1}{23.21}\right)} \approx 8.76°$$

For Point C (Electric Field Magnitude)

Since point C is inside the sphere, the electric field magnitude is zero: $$E_C = 0 \mathrm{N/C}$$ a) The electric field at point A is \(4.84 \times 10^3 \mathrm{N/C}\). b) The angle between the electric field and x-axis is \(\approx 8.76°\). c) The electric field at point C is \(0 \mathrm{N/C}\).

Key concepts

Gauss's Law

Understanding Gauss's Law is crucial when investigating electric fields created by charge distributions. In essence, Gauss's Law relates the electric field emanating from a closed surface to the charge enclosed within that surface. It states that the total electric flux through a closed surface, which is a concept meaning how much field is passing through, is equal to the charge enclosed divided by the permittivity of the medium (\(\epsilon_0\)). The mathematical expression for Gauss's Law is \(\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}\), where \(\Phi_E\) is the electric flux, \(Q_{\text{enc}}\) is the enclosed charge, and \(\epsilon_0\) is the permittivity of free space.

For a conductor like our solid sphere, Gauss's Law tells us something really intriguing: the electric field inside a conducting material in electrostatic equilibrium is zero. This is because the free charges will move in response to electric fields and distribute themselves on the surface until the field inside cancels out. Another consequence of Gauss's Law for a charged sphere is that the electric field outside it acts as if all the charge were concentrated at the center. This is why the calculations for the electric field outside the sphere in the exercise made use of this guideline and treated the charge as if it were a point charge at the center of the sphere.

Electric Field Magnitude

The electric field magnitude reflects the force a charge would experience in the vicinity of other charges. The field has both magnitude and direction, making it a vector quantity. For point charges and spherical charge distributions, as seen in our problem, the magnitude of the electric field at a distance \(r\) from the charge is given by the equation \(E = \frac{kQ}{r^2}\), where \(E\) is the magnitude of the electric field, \(k\) is Coulomb's constant (\(8.99 \times 10^9 \frac{\text{N m}^2}{\text{C}^2}\)), and \(Q\) is the charge creating the field.

For the point outside the sphere (Point A), observe that the formula takes into account only the distance from the surface and the charge. The situation becomes noticeably different inside the conductor (Point C), where the electric field magnitude is zero regardless of the position. This is a direct result of the characteristics of conductors and charge distribution under electrostatic conditions, which can be predicted based on Gauss's Law and the principle of superposition.

Electric Charge Distribution

Electric charge distribution is how charges are spread out in a material or space. In the context of our conducting sphere, the charges are free to move and will do so until they reach a state where the repulsive forces between them are balanced. In the end, all of the charges reside on the outer surface, creating a uniform surface charge distribution on a spherical conductor. This distribution has a significant effect on the electric field both inside and outside the sphere.

Inside the sphere, there is no electric field because the charges have arranged themselves such that their net effect cancels out any field within the conductor, as mentioned previously. This can be quite unexpected as intuitively one might think the charges throughout the sphere would contribute to the electric field throughout the volume. However, Gauss's Law, leveraging symmetry, shows us that within a uniformly charged spherical shell, or for any point within a solid sphere of charge, the net electric field is indeed zero, which aligns with the exercise's solution.