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Chapter 23: Problem 73

(a) Write the electrostatic potential energy of a pair of oppositely charged, closely spaced parallel plates as a function of their separation \(x,\) their area \(A,\) and the charge magnitude \(Q .\) (b) Differentiate with respect to \(x\) to find the magnitude of the attractive force between the plates. Why isn't the force equal to the charge on one plate times the electric field between the plates?

Short Answer

Expert verified
The electrostatic potential energy of the plates is \(U = \frac{Q^2 x}{2A\varepsilon_0}\). Differentiating with respect to \(x\) gives the attracting force as \(F = -\frac{Q^2}{2A\varepsilon_0}\). The force does not equal \(QE\) because this rule applies to point charges while a parallel plate is not a point charge.

Step by step solution

01

Write the electrostatic potential energy

The electrostatic potential energy \(U\) of a charged system is given by \(U = \frac{1}{2} QV\), where \(Q\) is the charge and \(V\) is the potential difference. The potential difference between the plates is \(V=E*x\) where \(E\) is the electric field and \(x\) is the separation between the plates. The electric field \(E\) is \(\frac{Q}{A\varepsilon_0}\), where \(A\) is the area of the plate and \(\varepsilon_0\) is the permittivity of free space. Substituting these into the equation for \(U\) gives \(U = \frac{1}{2} * Q * \frac{Qx}{A\varepsilon_0} = \frac{Q^2 x}{2A\varepsilon_0}\)
02

Differentiate to find the attracting force

The force \(F\) is the negative derivative of the potential energy with respect to the separation \(x\), i.e. \(F = -\frac{dU}{dx}\). Using the expression for \(U\) from step 1 and differentiating gives \(F= -\frac{d}{dx} (\frac{Q^2 x}{2A\varepsilon_0}) = -\frac{Q^2}{2A\varepsilon_0}\)
03

Explain the discrepancy

The force might be expected to be equal to the charge times the electric field, i.e. \(F = QE\). However, this is not the case here because this rule applies to point charges, and a parallel plate is not a point charge. This force is distributed over an area instead of being at a single point, so the actual force is less than \(QE\).

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Most popular questions from this chapter

A capacitor stores \(40 \mathrm{mJ}\) of energy when charged to \(100 \mathrm{V}\) (a) How much would it store at \(25 \mathrm{V} ?\) (b) What's its capacitance?

A cubical region \(1.0 \mathrm{m}\) on a side is located between \(x=0\) and \(x=1 \mathrm{m} .\) The region contains an electric field whose magnitude varies with \(x\) but is independent of \(y\) and \(z: E=E_{0}\left(x / x_{0}\right),\) where \(E_{0}=24 \mathrm{kV} / \mathrm{m}\) and \(x_{0}=6.0 \mathrm{m} .\) Find the total energy in the region.

What voltage is needed to put \(1.6 \mathrm{mC}\) on a \(100-\mu \mathrm{F}\) capacitor?

As an electrical engineer, you're asked to specify a capacitor that can store 12 mJ of energy. The largest capacitor that will physically fit on your circuit board is \(10 \mu \mathrm{F}\). The manufacturer produces capacitors with voltage ratings in multiples of \(25 \mathrm{V}\). What voltage do you specify?

A parallel-plate capacitor has plates with area \(50 \mathrm{cm}^{2}\) separated by \(25 \mu \mathrm{m}\) of polyethylene. Find its (a) capacitance and (b) working voltage.
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