Problem 90 - University Physics with Modern Physics

Question

Two very large horizontal sheets are 4.25 cm apart and carry equal but opposite uniform surface charge densities of magnitude \(\sigma\). You want to use these sheets to hold stationary in the region between them an oil droplet of mass 486 \(\mu\)g that carries an excess of five electrons. Assuming that the drop is in vacuum, (a) which way should the electric field between the plates point, and (b) what should \(\sigma\) be?

Step-by-Step Solution

Answer

(a) The electric field should point downward. (b) The required surface charge density \( \sigma \) is approximately \( 5.33 \times 10^{-10} \text{ C/m}^2 \).

Step 1: Analyze the Forces on the Oil Droplet

To keep the oil droplet stationary between the plates, the electric force must balance the gravitational force acting on it. The gravitational force on the droplet can be calculated by the formula: \\[ F_g = m \cdot g \] \where \( m = 486 \times 10^{-9} \text{ kg} \) (converted from micrograms) and \( g = 9.8 \text{ m/s}^2 \).

Step 2: Determine the Charge on the Oil Droplet

The charge of the oil droplet is determined by the excess electrons it carries. Since one electron has a charge of \( e = 1.6 \times 10^{-19} \text{ C} \), the total charge \( q \) on the oil droplet is: \\[ q = 5 \times (-1.6 \times 10^{-19}) \text{ C} \] \This results in a net negative charge for the droplet.

Step 3: Calculate the Required Electric Field Direction

To balance the negative charge of the droplet, the electric field must exert an upward force to counteract gravity. Since an electric field exerts force opposite to the field direction on a negative charge, the electric field must point downward.

Step 4: Calculate the Magnitude of the Electric Field

Use the balance of forces: \\[ F_e = F_g \] \where \( F_e = qE \) and \( F_g = 4.7628 \times 10^{-6} \text{ N} \). So, \( qE = mg \) becomes: \\[ E = \frac{mg}{q} \] \Plug in the values to calculate \( E \).

Step 5: Relate Electric Field to Surface Charge Density

Use the relation between the electric field and surface charge density in parallel plates: \\[ E = \frac{\sigma}{\varepsilon_0} \] \Solve for \( \sigma \): \\[ \sigma = E \cdot \varepsilon_0 \] \Substitute \( \varepsilon_0 = 8.854 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \).

Step 6: Solve for Surface Charge Density

Using the expression from Step 5, compute \( \sigma \) by substituting the calculated value of \( E \) from Step 4: \\[ \sigma = E \cdot 8.854 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \] \Perform the multiplication to find \( \sigma \).

Key Concepts

Surface Charge Density

Surface charge density, denoted by \(\sigma\), refers to the measure of electric charge per unit area on a surface. It plays a key role in determining the strength of the electric field due to the surface. For parallel plate capacitors or other systems with uniform surface charge, the electric field \(E\) related to surface charge density can be found using the relation:
\[E = \frac{\sigma}{\varepsilon_0}\]where \(\varepsilon_0\) is the vacuum permittivity, a constant approximately equal to \(8.854 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2\). This equation shows how \(\sigma\) affects the strength of the electric field between the charged plates. In practical scenarios like holding an oil droplet between charged plates, understanding surface charge density helps in calculating the electric force needed to balance other forces acting on the droplet, such as gravity.

Gravitational Force

Gravitational force \(F_g\) is the force exerted by the earth on an object due to its mass. It is calculated using the equation:
\[F_g = m \cdot g\]where \(m\) represents the mass of the object, and \(g\) is the acceleration due to gravity, usually taken as \(9.8 \text{ m/s}^2\). For instance, in the problem with the oil droplet, the mass is provided as \(486 \mu g\), which converts to \(486 \times 10^{-9} \text{ kg}\). Thus, the gravitational force acting downward on the droplet can be computed to ensure correct balance with other forces.
Gravitational force is always directed towards the center of the earth, thus contributing to the understanding of how to orient and what magnitude opposing forces like electric forces should have to achieve equilibrium.

Electric Force

Electric force \(F_e\) is the force exerted on a charged object within an electric field. It is given by the equation:
\[F_e = qE\]where \(q\) is the charge on the object and \(E\) is the electric field. If the charge is negative (like an oil droplet with excess electrons), the direction of the force is opposite to that of the electric field.
The balance of electric and gravitational forces is key when holding an object stationary between charged plates. The electric force must counterbalance the gravitational pull for equilibrium:
\[qE = mg\]Understanding this balance helps in determining both the direction of the electric field required (it must be opposite the gravitational force for negatively charged droplets) and its necessary strength.

Charge of an Electron

The charge of an electron is a fundamental property of matter, expressed as \(-1.6 \times 10^{-19} \text{ C}\). Electrons carry this standard charge, and their excess or deficit on an object leads to net positive or negative charges respectively.
In electrostatic calculations, such as with the oil droplet in this exercise, knowing the individual charge of an electron helps calculate the total charge \(q\) of the object:

• Multiply the number of excess electrons by the charge of one electron.

For example, with five excess electrons, the charge becomes:
\[q = 5 \times (-1.6 \times 10^{-19}) \text{ C}\]Knowing this value aids in determining the electric force required to balance out gravitational force and ensures the stationary condition of charged droplets in electric fields.