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Tutorial questions focusing upon the Coulomb field, electrostatic potentials and parallel plate capacitors
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\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, and take the relative permittivity of air to be 1.0005.
\nIf you got the wrong answer, did you remember to
\nThe $E$-field due to a point charge is
\n$\\displaystyle{q\\over 4 \\pi \\varepsilon_0 \\varepsilon_r r^2}$
\nradially away from the point charge. The force experienced in an electric field is generally $\\vec{F}=q\\vec{E}$, so the force on a point charge, $q_1$ due to another point charge, $q_2$ has a magnitude of
\n$\\displaystyle q_1 {q_2\\over 4 \\pi \\varepsilon_0 \\varepsilon_r r^2}$.
\nSince typically $\\varepsilon_r\\ge1$, any material between the two charges tends to reduce the force relative to the point charges in vacuum. It is important to note that the question asks for the percentage by which the force is reduced, not the percentage of the original force that remains.
\nThe force due to gravity is $mg$ towards the centre of the Earth, so one can calculate the force due the electric field and work out the mass that when mulitplied by $g$ has the same magnitude.
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\n$|\\vec{F}| =$ [[0]] Newtons.
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\nThe force is reduced by [[0]] %
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\nThe mass balanced by the electric field force is [[0]] kg
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\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 Coulomb field is for a point charge (the equation does not generally apply to all charge arrangements):
\nSymbol | \nDescription | \nUnits | \n
$E$ | \nElectric field strength | \nV/m (or N/C) | \n
$q$ | \nPoint charge | \nC | \n
$\\varepsilon_0$ | \nPermittivity of free space | \nF/m | \n
$\\varepsilon_r$ | \nRelative permittivity of medium betweem the point charge and where the field is being established | \nnone | \n
$r$ | \nDistance from the point charge - it's definitely not a radius! | \nm | \n
$\\hat{r}$ | \nThis is the unit radial vector. It is not a displacement (which would have units of m). | \nnone | \n
In the calculation of the magnitude of the electric field strength, care has to be taken to correctly account for units.
\nFrom the electric field, a force can be obtained using $F=qE$. For this problem the field has already been obtained, and all that's required is to multiply this by the magnitude of the second point charge.
\nSince the charges are opposite in sign, they attract.
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\n$\\displaystyle \\vec{E}= {q\\over 4 \\pi \\varepsilon_0 \\varepsilon_r r^2}\\hat{r}$.
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\n$q$ is [[2]] with units [[3]]
\n$\\varepsilon_0$ is [[4]] with units [[5]]
\n$\\varepsilon_r$ is [[6]] with units [[7]]
\n$r$ is [[8]] with units [[9]]
\n$\\hat{r}$ is [[10]] with units [[11]]
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\n$|\\vec{E}|=$ [[0]] [[1]]
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\n$F=$[[0]]
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\n$|\\vec{F}|=$ [[0]] [[1]]
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\n[[0]]
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$\\displaystyle{q\\over4\\pi\\varepsilon_0r^2}$.
When 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.
You might have calculated the force on the second point charge using the full Coulomb force formula,
\n$\\displaystyle F={q_1q_2\\over4\\pi\\varepsilon r^2}$
\nIf you approached the question this way you have missed a functional use of the electric field as it provides directly the potential for a force to exist, so that $\\vec{F}=q\\vec{E}$.
\nCommon errors include unit slips and forgetting to square the distance.
", "rulesets": {}, "variables": {"r1": {"name": "r1", "group": "Ungrouped variables", "definition": "random(1..10#1)", "description": "Random distance between 1 and 10 mm in steps of 1 mm (unit are mm)
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\n$|\\vec{E}|=$ [[0]] [[1]].
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\n$|\\vec{F}|=$ [[0]] [[1]].
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This question requires unit conversion, numerical calculations and some critical evaluation.
", "licence": "All rights reserved"}, "statement": "A point charge is placed at the origin of a Cartesian co-ordinate system, in vacuum. In this problem we shall use the standard boundary condition for the electrostatic potential from a point charge, i.e. $V(\\infty)=0\\,\\text{V}$.
\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": "Quantiative errors may occur due to unit conversion errors.
\nA common error is mis-remembering the Coulomb potential which is an inverse distance rule, and not an inverse-square (which is what we see for the electric field in Coulomb's Law). If you squared the distance, revise this section of the course.
\nThe correct formula for the electrostatic potential due to a point charge, taking the boudary condition that $V(\\infty)=0$, is
\n$\\displaystyle V(r)=\\frac{q}{4\\pi\\varepsilon r}$
\nwhere $r$ is the distance from the point charge. Note, the electrostatic potential is a scalar field - it does not have a direction.
\nIf you calculated the distance and potential at point $c$, then you may have not understood the symmetry of the system and the idea that a point charge has spherical equipotentials. Since the distance from the point charge to the points $a$ and $c$ are the same, so is the electrostatic potential.
\nThe potential difference between point $c$ and 'infinity' is obtainable directly from the definition of the potential difference, which is the energy per unit charge. Therefore the energy gained (they are both positive charges so therefore repel) is {q2}×10−6×$V_c$.
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\n$r_a=$ [[1]] cm.
\n$V(r_a)=$[[0]] Volts.
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\n$r_b=$ [[1]] cm.
\n$V(r_b)=$[[0]] Volts.
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\n$r_c=$ [[1]] cm.
\n$V(r_c)=$[[0]] Volts.
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\nEnergy gained = [[0]] J
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\n\nA point-charge is located at a point P in vacuum as shown on the schematic, indicative diagram (axes and points not to scale).
", "advice": "The electrostatic potential from a point charge is stated in the question. We have to take care in defining the meaning of the symbols and in using consistent units. In the question the \"normal\" units are not given as options, but in all cases the correct units can be derived from one or more of the options provided. For example, since 1A=1C/s, Coulombs are equivalent to Amp.seconds, and A.s/C is a dimensionless object (the units cancel).
\nFor the potential due to a point charge, we can determine the potential difference between the two points, A and B, due to a charge $q_1$ at P as
\n$\\displaystyle \\Delta V={q_1\\over4\\pi\\varepsilon}\\left({1\\over|\\vec{PA}|}-{1\\over|\\vec{PB}|}\\right)$,
\nwhere $\\vec{PA}$ is the vector from P to A and $\\vec{PB}$ is the vector from P to B.
\nWe do not know the value of $q_1$ -- this is to be found -- but we do know the energy change in moving an electron (with a charge of $-1.6\\times 10^{-19}$C) from A to B. This is related to the potential difference simply as $-e\\Delta V$. (Checking units, the p.d. is in Volt=J/C so multiplication by a charge yeilds an energy.) We're told the energy difference in eV, so we need to know how to convert between eV and Joules: 1 eV=$e$ J.
\nSo to obtain the charge at P, we need to solve
\n$\\displaystyle \\Delta \\text{P.E.}=-e\\Delta V=-{eq_1\\over4\\pi\\varepsilon}\\left({1\\over|\\vec{PA}|}-{1\\over|\\vec{PB}|}\\right)$
\nfor $q_1$ (it's the only unknown).
\nIf the charge at P is positive, then it will cost energy to move an electron from A to B, whereas if the charge at P is negative, we will gain energy in the move so the cost will be negative.
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\n$\\displaystyle V={q\\over4\\pi\\varepsilon r}$.
\nMatch the symbols to their descriptions. Some descriptions will be marked as partially correct (1 mark) but in all cases there is at least one option for full marks (2 marks). Only one option per symbol is allowed.
\n[[0]]
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\n$q$: [[0]]
\n$r$: [[1]]
\n$\\varepsilon_0$: [[2]]
\n$\\varepsilon_r$: [[3]]
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\n$V(r_a)-V(r_b)=$[[0]]
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\nChange in energy$=$[[0]]
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\nCharge at P is [[0]] nC.
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When 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.
The equations required to anwer this question are:
\n$\\displaystyle C={Q\\over V}={\\varepsilon A\\over d}$
\n$\\displaystyle E={V\\over d}={\\sigma \\over \\varepsilon}$
\n$\\displaystyle \\sigma = {Q\\over A}$
\nand
\nStored energy $\\displaystyle = {1\\over 2}QV$
\nwhere $C$ is the capacitance, $V$ is the potential difference between the plates, $A$ is the plate area, $d$ is the plate separation, $sigma$ is the charge density on a plate, $Q$ is the total charge on a plate and $\\varepsilon=\\varepsilon_0\\varepsilon_r$ is the permittivity of the material between the plates.
\nThe formula for the electric field in terms of the charge density and the permittivity is derived from the application of Gauss' Law to an infinite sheet of uniform charge density, so is only an approximation. It is pretty accurate between the plates away from the edge if the gap is small compared to the length-scale of the plate. At the edge of the PPC the field is dipolar (not uniform). If the plates are far apart compared to the diameter of the plate, then the approximation is poor.
\nFinally, there are requirements to convert between units, so if you got the right numbers except for where the decimal point is, then check your unit conversion.
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\n$|\\vec{E}|=$ [[1]]
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\n$|\\vec{E}|=$ [[0]] V/m
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\n$\\sigma=$ [[1]]
\nCalculate the charge density on the positive plate.
\n$\\sigma=$ [[0]] C.m$^{-2}$
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\n$C=$ [[1]]
\nCalculate the capacitance.
\n$C=$ [[0]] Farads
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\n$\\varepsilon_r=$ [[1]]
\nCalculate the value of the relative permittivity.
\n$\\varepsilon_r=$ [[0]]
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\n$Q_{\\rm max}=$ [[0]] Coulombs
\nHow much energy is required to increase the charge from {qn} nC to this maximum value?
\nIncrease in energy = [[1]] nano-Joules.
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