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Two \(2.00 \mathrm{cm} \times 2.00 \mathrm{cm}\) plates that form a parallel- plate capacitor are charged to ±0.708 nC. What are the electric field strength inside and the potential difference across the capacitor if the spacing between the plates is (a) \(1.00 \mathrm{mm}\) and (b) \(2.00 \mathrm{mm} ?\)

Short Answer

Expert verified
The electric field strength inside the capacitor is \(2.00 × 10^{5} N/C\). The potential difference across the capacitor is 200 V for a spacing of 1.00 mm and 400 V for a spacing of 2.00 mm.

Step by step solution

01

Calculate the charge density

The charge density (σ) on the surface of the capacitor can be calculated using the formula: \(σ = Q/A\), where Q is the charge on the plates and A is the area of the plates. Our charge Q is 0.708 nC (convert this to C by multiplying with \(10^{-9}\)) and area A is 4.00 \(cm^{2}\) (convert this to \(m^{2}\) by multiplying with \(10^{-4}\)). Calculating, we find \(σ = 0.708 × 10^{-9} / 4.00 × 10^{-4} = 1.77 × 10^{-6} C/m^{2}\).
02

Calculate the electric field strength

The electric field strength (E) inside a capacitor can be found with the formula: \(E = σ / ε_0\), where \(ε_0\) is the permittivity of free space (\(ε_0 = 8.85 × 10^{-12} C^2/N*m^2\)). Thus, \(E = 1.77 × 10^{-6} / 8.85 × 10^{-12} = 2.00 × 10^{5} N/C\).
03

Calculate the potential difference: Case (a)

The potential difference (V) across the capacitor can be found with the formula: \(V = E * d\), where d is the distance between the plates. For case (a), d is \(1.00 mm\) (convert this to m by multiplying with \(10^{-3}\)). Thus, \(V = 2.00 × 10^{5} * 1 × 10^{-3} = 200 V\).
04

Calculate the potential difference: Case (b)

For case (b), the distance d is \(2.00 mm\), also in meters. Thus, \(V = 2.00 × 10^{5} * 2 × 10^{-3} = 400 V\).

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Key Concepts

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

Electric Field Strength
The electric field strength is a fundamental concept in electromagnetism, describing the force per unit charge experienced by a stationary charge when placed in an electric field. It is denoted by the symbol E and measured in newtons per coulomb (N/C). In the context of a parallel-plate capacitor, the electric field is uniform between the plates and can be calculated as

\frac{\sigma}{\varepsilon_0}, where \(\sigma\) is the charge density on the plates, and \(\varepsilon_0\) is the permittivity of free space. The electric field is directed from the positive plate to the negative one and represents the strength and direction of the electrostatic force that would act on a positive test charge placed in the field. The higher the charge density or the lower the permittivity of free space, the greater the electric field strength.
Charge Density
Charge density, often symbolized by \(\sigma\), is a measure of the electric charge per unit area on a surface. In units, it is expressed in coulombs per square meter (C/m2). For a parallel-plate capacitor, the charge density can be calculated by dividing the total charge \(Q\) on one plate by the area \(A\) of that plate, leading to the formula

\(\sigma = \frac{Q}{A}\). It is critical to convert all measurements to SI units to ensure accuracy in calculations. The significance of charge density lies in its direct proportionality to the electric field strength and potential difference across the capacitor; a higher charge density produces a stronger electric field and a higher potential difference.
Potential Difference
The potential difference, commonly referred to as voltage (V), between two points in an electric field is a measure of the work done by the electric field to move a unit charge from one point to the other. In the case of a parallel-plate capacitor, this difference corresponds to the energy difference per charge between the two plates. The potential difference is determined using the formula

\(V = E \cdot d\), where \(E\) is the electric field strength, and \(d\) is the separation between the plates. The concept of potential difference is essential in understanding the energy storage of capacitors: the greater the potential difference for a given charge, the more energy is stored.
Permittivity of Free Space
The permittivity of free space, denoted by \(\varepsilon_0\), is a fundamental constant in physics that describes how strongly the electric field affects free space. Its value is approximately \(8.85 \times 10^{-12} C^2/N\cdot m^2\). This constant is used to relate the electric field strength within a vacuum to the charge density, following the equation

\(E = \frac{\sigma}{\varepsilon_0}\). The permittivity of free space also influences how electric fields interact with materials, as it is the baseline measure for determining the relative permittivity or dielectric constant of various substances. Understanding \(\varepsilon_0\) helps in comprehending how different materials influence the capacitance and electric field behavior in capacitive systems.

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Most popular questions from this chapter

An electric dipole has dipole moment \(p .\) If \(r \gg s,\) where \(s\) is the separation between the charges, show that the electric potential of the dipole can be written $$V=\frac{1}{4 \pi \epsilon_{0}} \frac{p \cos \theta}{r^{2}}$$ where \(r\) is the distance from the center of the dipole and \(\theta\) is the angle from the dipole axis.

What is the escape speed of an electron launched from the surface of a 1.0 -cm-diameter glass sphere that has been charged to \(10 \mathrm{nC} ?\)

The hydrogen molecular ion \(\mathrm{H}_{2}^{+}\), with one electron and two protons, is the simplest molecule. The equilibrium spacing between the protons is \(0.11 \mathrm{nm}\). Suppose the electron is at the midpoint between the protons and moving at \(1.5 \times 10^{6} \mathrm{m} / \mathrm{s}\) perpendicular to a line between the protons. How far (in nm) does the electron move before reaching a turning point? Because of their larger mass, the protons remain fixed during this interval of time. Note \(\Rightarrow\) An accurate description of \(\mathrm{H}_{2}^{+}\) requires quantum mechanics. Even so, a classical calculation like this provides some insight into the molecule.

Electrodes of area \(A\) are spaced distance \(d\) apart to form a parallel-plate capacitor. The electrodes are charged to \(\pm q\) a. What is the infinitesimal increase in electric potential energy \(d U\) if an infinitesimal amount of charge \(d q\) is moved from the negative electrode to the positive electrode? b. An uncharged capacitor can be charged to \(\pm Q\) by transferring charge \(d q\) over and over and over. Use your answer to part a to show that the potential energy of a capacitor charged to \(\pm Q\) is \(U_{\mathrm{cap}}=\frac{1}{2} Q \Delta V_{\mathrm{c}}\)

A proton is fired from far away toward the nucleus of a mercury atom. Mercury is element number 80 , and the diameter of the nucleus is \(14.0 \mathrm{fm}\). If the proton is fired at a speed of \(4.0 \times 10^{7} \mathrm{m} / \mathrm{s},\) what is its closest approach to the surface of the nucleus? Assume the nucleus remains at rest.

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