Question

If a charge is held in place above a large, flat, grounded, conducting slab, such as a floor, it will experience a downward force toward the floor. In fact, the electric field in the room above the floor will be exactly the same as that produced by the original charge plus a "mirror image" charge, equal in magnitude and opposite in sign, as far below the floor as the original charge is above it. Of course, there is no charge below the floor; the effect is produced by the surface charge distribution induced on the floor by the original charge. a) Describe or sketch the electric field lines in the room above the floor. b) If the original charge is \(1.00 \mu \mathrm{C}\) at a distance of \(50.0 \mathrm{~cm}\) above the floor, calculate the downward force on this charge. c) Find the electric field at (just above) the floor, as a function of the horizontal distance from the point on the floor directly under the original charge. Assume that the original charge is a point charge, \(+q,\) at a distance \(a\) above the floor. Ignore any effects of walls or ceiling. d) Find the surface charge distribution \(\sigma(\rho)\) induced on the floor. e) Calculate the total surface charge induced on the floor.

Short answer

Question: Describe the electric field lines in the room above the floor, calculate the downward force on the original charge, find the electric field just above the floor as a function of the horizontal distance, find the surface charge distribution induced on the floor, and calculate the total surface charge induced on the floor. Answer: a) The electric field lines in the room above the floor are symmetric with respect to the floor, starting from the positive charge and ending at the negative charge, similar to what we would observe when two opposite charges are close to each other. b) The downward force experienced by the original charge is 9.00 x 10^-3 N. c) The electric field just above the floor and at a distance ρ from the point directly below the original charge is given by: E_total = E_(q_1) + E_(q_2) d) The surface charge distribution on the floor can be determined by using the relation between electric field and surface charge density: σ(ρ) = ε₀ E_⊥ e) The total surface charge induced on the floor can be calculated by integrating the surface charge density over the entire floor: Q_total = ∫ σ(ρ) dA

Step-by-step solution

a) Electric field lines

The electric field lines in the room above the floor are produced by the original point charge and its mirror image charge. The field lines start from the positive charge and end at the negative charge, so they are symmetric with respect to the floor. This behavior is similar to what we observe when two opposite charges are close to each other.

b) Downward force on original charge

To find the downward force on the original charge, we need to calculate the electric field produced by the mirror image charge on the original charge. Let the original charge be \(+q\). By symmetry, the mirror image charge will be \(-q\) and the distance between them will be \(2a\), where \(a\) is the distance of the original charge from the floor. The electric field produced by the mirror image charge, \(E\), is given by: \(E = \frac{kq}{(2a)^2}\) The downward force experienced by the original charge, \(F\), is: \(F = qE = \frac{q^2 k}{(2a)^2}\) We are given that \(q = 1.00 \mu \mathrm{C}\) and \(a = 50.0 \mathrm{~cm}\). Using these values and the constant \(k = 9 \times 10^9 \mathrm{N m^2C^{-2}}\), we can calculate the downward force: \(F = \frac{(1.00 \times 10^{-6})^2 \times 9 \times 10^9}{(2 \times 0.50)^2}\) \(F = 9.00 \times 10^{-3} \mathrm{N}\)

c) Electric field at the floor

Let the horizontal distance from the point on the floor directly under the original charge be \(\rho\). To find the electric field at the floor as a function of distance \(\rho\), we can use the principle of superposition. The electric field at this point is due to both the original charge \(+q\) and the mirror image charge \(-q\). For a point charge, the radial electric field is given by: \(E_q = \frac{k |q|}{r^2}\) where \(r\) is the distance from the charge. We need to find the individual electric fields due to the original and image charges, and then add them vectorially. The distance from the original charge to the point \((\rho, 0)\) on the floor is: \(r_1 = \sqrt{a^2 + \rho^2}\) The distance from the image charge to the point \((\rho, 0)\) on the floor is: \(r_2 = \sqrt{a^2 + \rho^2}\) Now, let's find the electric field vectors due to each charge: \(E_{q_1} = \frac{kq}{r_1^2}\) \(E_{q_2} = \frac{-kq}{r_2^2}\) To find the total electric field at the point \((\rho, 0)\), we add the electric field vectors: \(E_{total} = E_{q_1} + E_{q_2}\) The electric field will be just above the floor and at a distance \(\rho\) from the point directly below the original charge.

d) Surface charge distribution on the floor

To find the surface charge distribution, \(\sigma(\rho)\), induced on the floor, we can use the relation between electric field and surface charge density: \(\sigma(\rho) = \epsilon_0 E_\perp\) where \(E_\perp\) is the electric field perpendicular to the floor, and \(\epsilon_0\) is the permittivity of free space. To find \(E_\perp\), we can decompose the total electric field into horizontal and vertical components. We will then use the vertical component, since it is perpendicular to the floor: \(E_\perp = E_{total} \sin{\theta}\) where \(\theta\) is the angle between the total electric field and the horizontal axis. Finally, using the total electric field calculated in part (c), and the angle, we can find the surface charge density: \(\sigma(\rho) = \epsilon_0 E_\perp\)

e) Total surface charge induced on the floor

To calculate the total surface charge induced on the floor, we need to integrate the surface charge density over the entire floor: \(Q_{total} = \int \sigma(\rho) dA\) where \(dA\) is the differential area element of the floor. Since the floor is a flat, conducting slab, we can use polar coordinates to integrate the surface charge density: \(dA = \rho d\rho d\phi\) Integrating \(\sigma(\rho)\), found in part (d), over the entire floor, gives us the total induced surface charge.