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This question focuses upon the permittivity of materials in parallel plate capacitors.
\nA parallel plate capacitor has circular plates of radius {radiusmm} mm and the distance between the plates, $d$, is {distanceu} μm. The capacitor is connected to a battery indicated as {battery} V.
\nDiscs that fill the space between the plates comprised from various materials can be placed between the electrodes and the energy stored recorded.
\nWhen providing numerical answers you may express them using scientific notation. Express values to four significant figures and use the values of physical constants as provided in the course notes.
", "advice": "The first step is to note that the energy stored is
\n$\\displaystyle {CV^2\\over2},$
\nand that
\n$\\displaystyle C={\\varepsilon A\\over d},$
\nwhere $C$ is the capacitance (F), $V$ is the potential difference between the two plates (volts), $\\varepsilon$ is the permittivity (F/m) of the medium between the plates of area $A$ (m$^2$) and separation $d$ (m).
\nThe relative permittivity of the medium is obtainable:
\n$\\displaystyle {\\rm energy~stored}={1\\over 2}CV^2={1\\over2}{\\varepsilon_0\\varepsilon_r A\\over d}V^2\\Rightarrow \\varepsilon_r={2 d \\times {\\rm stored~energy}\\over \\varepsilon_0 AV^2}$.
\nWe have all the quantities so $\\varepsilon_r$ can be obtained.
\nNoting that the relative permittivity is inversely proportional to the voltage (squared), if the voltage of the battery is lower than indicated then the permittivity is higher than previously estimated, and so therefore is the capacitance.
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\nEnergy stored = [[0]]
\n$C$ is the [[1]] with units [[2]]
\n$V$ is the [[3]] with units [[4]]
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\n$\\varepsilon_r(A)=$ [[0]]
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\nDoes the error in the indicated value of the voltage of the battery mean that the value of the permittivity of material A previously is an under- or over-estimate?
\n[[0]]
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