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CP Two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by \(2.20 \mathrm{~cm} .\) (a) If the surface charge density for each plate has magnitude \(47.0 \mathrm{nC} / \mathrm{m}^{2}\), what is the magnitude of \(\overrightarrow{\boldsymbol{E}}\) in the region between the plates? (b) What is the potential difference between the two plates? (c) If the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference?

Short Answer

Expert verified
(a) 5316 N/C, (b) 117 V, (c) Field remains 5316 N/C, potential difference doubles to 235 V.

Step by step solution

01

Determine the Electric Field between the Plates (a)

The electric field (\overrightarrow{E}) between two large, parallel plates with surface charge density \sigma can be calculated using the formula: \[ E = \frac{\sigma}{\varepsilon_0} \] where \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{~C}^{2}/\mathrm{N} \cdot \mathrm{m}^{2} is the permittivity of free space. Plug in the given value: \[ \sigma = 47.0 \times 10^{-9} \mathrm{~C/m}^{2} \] \[ E = \frac{47.0 \times 10^{-9}}{8.85 \times 10^{-12}} = 5316.38 \mathrm{~N/C} \] Therefore, the magnitude of the electric field E is approximately \(5316 \mathrm{~N/C}\).
02

Calculate Potential Difference between Plates (b)

The potential difference V between the plates is given by: \[ V = E \cdot d \] where \(d = 2.20 \mathrm{~cm} = 0.022 \mathrm{~m}\). Use the electric field value calculated in step 1: \[ V = 5316 \times 0.022 = 116.96 \mathrm{~V} \] Thus, the potential difference between the plates is approximately \(117 \mathrm{~V}\).
03

Analyze the Effects of Doubling the Separation (c)

If the separation between the plates is doubled, it does not change the electric field \overrightarrow{E} because the surface charge density \(\sigma\) remains the same, and \(E = \frac{\sigma}{\varepsilon_0}\) is independent of the plate separation. Therefore, \(E\) remains at \(5316 \, \mathrm{N/C} \). For the potential difference: \[ V' = E \times (2d) = 5316 \times 0.044 = 234.72 \mathrm{~V} \] Thus, when the separation doubles, the potential difference also doubles to approximately \(235 \mathrm{~V}\).

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Key Concepts

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

Surface Charge Density
Surface charge density, denoted by \( \sigma \), is a measure of the amount of electric charge per unit area on a surface of a conductor. In our context, it plays a pivotal role in determining the electric field between two parallel plates. The charge is distributed evenly over the surface, and this uniform distribution is what we refer to as surface charge density.
For two large, parallel conducting plates with opposite charges, the electric field \( E \) between them can be directly related to surface charge density through the equation:
  • \( E = \frac{\sigma}{\varepsilon_0} \)
This shows that the electric field strength is constant between the plates and directly proportional to the surface charge density. Therefore, higher the surface charge density, stronger will be the electric field between the plates. In our scenario, with \( \sigma = 47.0 \times 10^{-9} \mathrm{~C/m}^{2} \), the calculated electric field is \( 5316 \mathrm{~N/C} \), providing an example of how surface charge density influences the electric field.
Potential Difference
Potential difference, commonly referred to as voltage (\( V \)), between two points in an electric field is the work done to move a unit charge from one point to another. For parallel plates, the potential difference can be determined using the electric field and the distance separated by the two plates:
  • \( V = E \cdot d \)
Here, \( d \) is the separation distance between the plates, which is \( 0.022 \) meters in this instance. By knowing \( E \), which we've calculated to be \( 5316 \mathrm{~N/C} \), the potential difference between the plates becomes \( 117 \mathrm{~V} \).
This demonstrates that the potential difference is directly proportional to both the electric field and the distance between the plates. Additionally, if the distance doubles, like in our exercise step 3, the potential difference will also double. Such properties of potential difference are crucial for understanding energy storage in capacitors and many electronic applications.
Permittivity of Free Space
The permittivity of free space, symbolized as \( \varepsilon_0 \), is a constant that describes how electric fields interact with the vacuum (or free space). Its value is approximately \(8.85 \times 10^{-12} \mathrm{~C}^{2}/\mathrm{N} \cdot \mathrm{m}^{2}\).

This constant is essential when calculating the electric field between surfaces like our parallel plates configuration. Using the surface charge density \( \sigma \) and the permittivity of free space \( \varepsilon_0 \), the electric field \( E \) is given by:
  • \( E = \frac{\sigma}{\varepsilon_0} \)
This equation highlights the role of \( \varepsilon_0 \) in moderating the effect of surface charge density on the electric field. As a fundamental physical constant, \( \varepsilon_0 \) serves to bridge the concepts of electric field, charge, and geometric arrangement of conductors, underscoring its significance in electromagnetic theory.

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Most popular questions from this chapter

A point charge \(q_{1}\) is held stationary at the origin. A second charge \(q_{2}\) is placed at point \(a\), and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \mathrm{~J}\). When the second charge is moved to point \(b\), the electric force on the charge does \(-1.9 \times 10^{-8} \mathrm{~J}\) of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b\) ?

(a) How much work would it take to push two protons very slowly from a separation of \(2.00 \times 10^{-10} \mathrm{~m}\) (a typical atomic distance) to \(3.00 \times 10^{-15} \mathrm{~m}\) (a typical nuclear distance)? (b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation?

A point charge \(q_{1}=+5.00 \mu \mathrm{C}\) is held fixed in space. From a horizontal distance of \(6.00 \mathrm{~cm}\), a small sphere with mass \(4.00 \times 10^{-3} \mathrm{~kg}\) and charge \(q_{2}=+2.00 \mu \mathrm{C}\) is fired toward the fixed charge with an initial speed of \(40.0 \mathrm{~m} / \mathrm{s}\). Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is \(25.0 \mathrm{~m} / \mathrm{s}\) ?

Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere \(A\) has a radius three times that of sphere \(B\). Let \(Q_{A}\) and \(Q_{B}\) be the charges on the two spheres, and let \(E_{A}\) and \(E_{B}\) be the electric-field magnitudes at the surfaces of the two spheres. What are (a) the ratio \(Q_{B} / Q_{A}\) and (b) the ratio \(E_{B} / E_{A} ?\)

A very long cylinder of radius \(2.00 \mathrm{~cm}\) carries a uniform charge density of \(1.50 \mathrm{nC} / \mathrm{m}\). (a) Describe the shape of the equipotential surfaces for this cylinder. (b) Taking the reference level for the zero of potential to be the surface of the cylinder, having potentials of \(10 \mathrm{~V}, 20 \mathrm{~V}\), and \(30 \mathrm{~V}\). are the equipotential surfaces equally spaced? If not, do they get closer together or farther apart as potential increases?

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