Question

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by 45.0 mm, and the potential difference between them is 360 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 nC? (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.

Short answer

The electric field is 8000 V/m, the force on the charge is \(1.92 \times 10^{-5}\) N, and the work done is \(8.64 \times 10^{-7}\) J.

Step-by-step solution

Calculate Electric Field

To find the magnitude of the electric field \( E \) between the plates, use the formula \( E = \frac{V}{d} \), where \( V \) is the potential difference and \( d \) is the separation distance between the plates. Here, \( V = 360 \, \text{V} \) and \( d = 45.0 \, \text{mm} = 0.045 \, \text{m} \). Thus, \( E = \frac{360}{0.045} = 8000 \, \text{V/m} \). The magnitude of the electric field is 8000 V/m.

Calculate the Force on the Charge

The force \( F \) exerted on a charge \( q \) by an electric field \( E \) is given by \( F = qE \). Here, \( q = 2.40 \, \text{nC} = 2.40 \times 10^{-9} \, \text{C} \) and \( E = 8000 \, \text{V/m} \). So, \( F = 2.40 \times 10^{-9} \times 8000 = 1.92 \times 10^{-5} \, \text{N} \). The force on the charge is \( 1.92 \times 10^{-5} \, \text{N} \).

Calculate the Work Done on the Charge

The work done \( W \) on a charge by a uniform electric field is given by \( W = qV \). With \( q = 2.40 \times 10^{-9} \, \text{C} \) and \( V = 360 \, \text{V} \), \( W = 2.40 \times 10^{-9} \times 360 = 8.64 \times 10^{-7} \, \text{J} \). The work done on the charge is \( 8.64 \times 10^{-7} \, \text{J} \).

Compare Work Done to Potential Energy Change

The change in electric potential energy \( \Delta U \) of the charge is given by \( \Delta U = qV \), which is identical to the work done \( W \) because work done is equal to the change in potential energy in a conservative force field. Thus, \( \Delta U = 8.64 \times 10^{-7} \, \text{J} \). This value matches the work done computed in Step 3, confirming the consistency.

Key concepts

Potential Difference

The concept of potential difference is crucial when studying electric fields and charges. Potential difference between two points in an electric field is the work done to move a unit charge from one point to the other. It is measured in volts (V). In our exercise, the potential difference between two parallel plates is 360 V. This potential difference creates an electric field that influences the movement of charges.

To calculate the electric field ( \( E \) ), use the formula:

• \( E = \frac{V}{d} \)

where \( V \) is the potential difference and \( d \) is the distance between the plates. For instance, with a 45.0 mm separation and a 360 V potential difference, the electric field amounts to 8000 V/m. This electric field dictates the force exerted on any charge placed between the plates.

Force on a Charge

In a uniform electric field, the force experienced by a charge is determined by the field's influence on the charge's properties. The force ( \( F \) ) on a charge is calculated by:

• \( F = qE \)

Here, \( q \) represents the magnitude of the charge, and \( E \) is the electric field strength. In our scenario, a charge of \( +2.40 \) nC placed in an electric field of 8000 V/m experiences a force of \( 1.92 \times 10^{-5} \) N.

This force is what causes the charge to move, illustrating fundamental principles of electromagnetism. Forces in electric fields are essential for explaining phenomena such as motion in circuits and the operation of various electrical devices.

Work Done

Work done in moving a charge within an electric field is a pivotal principle in electromagnetism. The work ( \( W \) ) performed by the electric field on a charge is given by:

• \( W = qV \)

where \( q \) is the charge and \( V \) is the potential difference. For this exercise, moving a charge of \( 2.40 \times 10^{-9} \) C across a potential difference of 360 V results in work done equal to \( 8.64 \times 10^{-7} \) J.

The concept of work done highlights the energy changes when a charge moves through an electric field. It acts as a bridge connecting electric potential and kinetic energy. The energy relationship in such scenarios allows us to design and understand electric circuits and energy storage systems.

Potential Energy Change

The change in potential energy ( \( \Delta U \) ) when a charge moves within an electric field is intrinsically linked to the work done. Potential energy change is calculated by:

• \( \Delta U = qV \)

In a conservative electric field, the potential energy change equals the work done. For our charge moving between plates with a potential difference of 360 V, \( \Delta U \) is \( 8.64 \times 10^{-7} \) J. This confirms the consistency with the work done computation.

Understanding potential energy changes helps in visualizing energy transfer in systems and guides the analysis of electric power distribution and usage. The principle that work done equals potential energy change is fundamental in energy conservation studies and electronics.