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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 \(+2.40 \mathrm{nC} ?\) (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

Expert verified
The magnitude of the electric field is \(8000 N/C\), the force on a \(+2.40 nC\) charge in the field is \(1.92*10^{-5}N\), and the work done by the field on the particle is \(8.64*10^{-7}J\), this is also the change in the potential energy of the same charge.

Step by step solution

01

Calculate the Electric Field

Calculate the magnitude of the electric field in the region between the plates, using the formula for the field strength of parallel plates: \(E = V/d\). Here, \(V = 360V\) is the potential difference and \(d = 45.0mm = 0.045m\). Substituting these values in the given formula gives: \(E = 360V / 0.045m = 8000 N/C\).
02

Calculate the Force on the Charge

Use the formula \(F = E * q\) to calculate the magnitude of the force this field exerts on a particle with charge \(q = +2.40 nC = 2.4*10^{-9}C\). Substituting these values into the formula yields: \(F = 8000N/C * 2.4*10^{-9}C = 1.92*10^{-5}N\).
03

Compute the Work Done by the Field

Use the formula for work done by an electric field on a charge: \(W = q * \Delta V\). Here, \(\Delta V = 360V\) is the voltage difference and \(q = 2.4*10^{-9}C\). Substituting these values into the formula yields: \(W = 2.4*10^{-9}C * 360V = 8.64*10^{-7}J\).
04

Compare Work Done and Change in Potential Energy

In an electric field, the work done on a charge is equal to the change in its potential energy. Hence, the calculated work done (\(8.64*10^{-7}J\)) is equal to the change in potential energy of the charge.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field Strength
Understanding the electric field strength is crucial when studying the effects of electric fields. It is a measure of the force a field will exert on a unit positive charge placed within it. Mathematically, electric field strength, symbolized by 'E', is given by the formula \(E = \frac{V}{d}\), where 'V' is the voltage (or electric potential difference) across the plates, and 'd' is the distance between them.

For parallel plates with a uniform field, this can be simplified since the field is considered consistent across the entire distance between the plates. In the example provided, calculating the strength given a potential difference of 360V and a separation of 0.045m leads us to find an electric field strength of 8000 N/C. It's significant to note this field strength indicates how strong the force per unit charge will be in the space between the plates.
Force on a Charge in an Electric Field
Once the electric field strength is known, we can calculate the force experienced by any charge placed within it. The force 'F' acting on a charge 'q' in an electric field of strength 'E' is defined by the equation \(F = E \cdot q\).

It's vital to ensure the charge is expressed in Coulombs when using this formula. For our example, the charge is a particle with a value of \(+2.40 \mathrm{nC}\), or \(2.4 \times 10^{-9} \mathrm{C}\), resulting in a force calculated as \(1.92 \times 10^{-5} \mathrm{N}\). This indicates how much push or pull the particle would experience from the electric field.
Work Done by Electric Field
The concept of work in physics relates to the amount of energy transferred when a force moves an object over a distance. For charges in electric fields, the work done by the field to move a charge is related to the charge's voltage difference while moving across the field. It's given by the equation \(W = q \cdot \Delta V\), where 'W' is the work, \(\Delta V\) is the voltage difference, and 'q' is the charge.

In our textbook case, when a charge moves from the higher-potential plate to the lower, the work done by the electric field on the particle is calculated to be \(8.64 \times 10^{-7} \mathrm{J}\). This work can be thought of as the energy imparted to the charge by the electric field as it moves through the potential difference.
Electric Potential and Potential Energy
Electric potential is a measure of the potential energy per unit charge at a certain location within a field and is influenced by the electric field strength and the position within the field. In terms of energy, potential energy represents the stored energy of a charge due to its position in an electric field. It can be calculated using the same equation for work, \(W = q \cdot \Delta V\), since work done on a charge by an electric field is equivalent to the change in the charge's potential energy.

To link this to our example, when we say a charge moves from a higher-potential plate to a lower one, we infer that its electric potential energy decreases. The calculated work in the example, \(8.64 \times 10^{-7} \mathrm{J}\), corresponds to the decrease in potential energy as the charge moves between the plates, showcasing the transformation of potential energy into kinetic or other forms of energy.

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Most popular questions from this chapter

A thin spherical shell with radius \(R_{1}=3.00 \mathrm{~cm}\) is concentric with a larger thin spherical shell with radius \(R_{2}=5.00 \mathrm{~cm}\). Both shells are made of insulating material. The smaller shell has charge \(q_{1}=+6.00 \mathrm{nC}\) distributed uniformly over its surface, and the larger shell has charge \(q_{2}=-9.00 \mathrm{nC}\) distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the fol- (ii) \(r=4.00 \mathrm{~cm}\) lowing distance from their common center: (i) \(r=0\) (iii) \(r=6.00 \mathrm{~cm} ?\) (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

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