Problem 39 - University Physics with Modern Physics

Question

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by \(45.0 \mathrm{mm},\) and the potential difference between them is 360 \(\mathrm{V}\) (a) What is the magnitude of the electric field (assumed to be uniform) in the region between the plates? (b) What is the magnitude of the force this field exerts on a particle with charge \(+2.40 \mathrm{nC}^{\prime}\) (c) Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower. (d) Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

Step-by-Step Solution

Answer

(a) 8000 V/m, (b) 1.92 × 10^{-5} N, (c) 8.64 × 10^{-7} J, (d) Same work and potential energy change.

Step 1: Calculate Electric Field (E)

The electric field between two parallel plates is given by the formula \( E = \frac{V}{d} \), where \( V \) is the potential difference and \( d \) is the separation distance. Given \( V = 360 \; \mathrm{V} \) and \( d = 45.0 \; \mathrm{mm} = 0.045 \; \mathrm{m} \), plugging these values into the formula gives \( E = \frac{360}{0.045} = 8000 \; \mathrm{V/m} \).

Step 2: Calculate Force (F) on the Charge

The force on a charge in an electric field is calculated using \( F = qE \), where \( q \) is the charge and \( E \) is the electric field. With \( q = +2.40 \; \mathrm{nC} = 2.40 \times 10^{-9} \; \mathrm{C} \) and \( E = 8000 \; \mathrm{V/m} \), the force is \( F = 2.40 \times 10^{-9} \times 8000 = 1.92 \times 10^{-5} \; \mathrm{N} \).

Step 3: Calculate Work Done (W) by the Field

The work done by the electric field when moving a charge over a potential difference is given by \( W = qV \). Using \( q = 2.40 \times 10^{-9} \; \mathrm{C} \) and \( V = 360 \; \mathrm{V} \), we find \( W = 2.40 \times 10^{-9} \times 360 = 8.64 \times 10^{-7} \; \mathrm{J} \).

Step 4: Compare Work Done and Potential Energy Change

The change in electric potential energy for a charge moving through a potential difference \( V \) is also given by \( \Delta U = qV \). As \( \Delta U = 8.64 \times 10^{-7} \; \mathrm{J} \), same as work done by the field, the work done by the electric field on the charge matches the change in potential energy.

Key Concepts

Electric Field Calculation

To determine the electric field between two parallel plates, we rely on a straightforward formula: \[ E = \frac{V}{d} \] where \( E \) signifies the electric field, \( V \) is the potential difference, and \( d \) is the distance separating the plates. It becomes simple when you recognize it as the ratio of potential voltage to separation distance. With the particulars from the problem, where \( V \) equals 360 volts and \( d \) is 0.045 meters (converting 45 mm into meters), it results in an electric field magnitude of \( 8000 \, \text{V/m} \).
This uniform field indicates a constant force experienced by any charge placed between these plates, facilitating consistent calculations for related quantities.

Force on a Charge in an Electric Field

The force that an electric field exerts on a charge is calculated by the formula: \[ F = qE \] where \( F \) is the force, \( q \) is the charge, and \( E \) is the electric field.

• Given a charge \( q = +2.40 \, \text{nC} = 2.40 \times 10^{-9} \, \text{C} \)
• Using the previously found electric field \( E = 8000 \, \text{V/m} \)

Plugging these values into the formula tells us the force exerted is \( 1.92 \times 10^{-5} \, \text{N} \).
This illustrates how electric fields can exert influence on charged particles, propelling them based on the field's intensity and the charge magnitude.

Work Done by Electric Field

When a charge moves in an electric field, the work done by that field is determined by \[ W = qV \] where \( W \) is the work done, \( q \) is the charge, and \( V \) is the potential difference. The process involves the charge moving from one potential to another, transferring energy.

• Substituting \( q = 2.40 \times 10^{-9} \, \text{C} \)
• And \( V = 360 \, \text{V} \)

Results in a calculated work amount of \( 8.64 \times 10^{-7} \, \text{J} \).
Work is essentially the energy required to move the charge across the plates, emphasizing the relationship between electric fields and energy transfer.

Potential Energy Change in Electric Field

The change in potential energy for a charge within an electric field is captured by the equation: \[ \Delta U = qV \] where \( \Delta U \) is the change in potential energy. In this exercise, it mirrors the work done formula, showing that the work done by the electric field exactly equals the change in the charge’s potential energy.
Using values

• \( q = 2.40 \times 10^{-9} \, \text{C} \)
• \( V = 360 \, \text{V} \)

results in a potential energy change of \( 8.64 \times 10^{-7} \, \text{J} \).
This correspondence illustrates the conservation of energy principles, reflecting how energy changes its form but remains constant in terms of quantity when moving through the electric field.