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ENG1003: Digital tutorial, Week 3, Semester 2

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This question is about the electric field a distance from a point charge in vacuum.  Coulomb's law states that the electric field strength from a point charge in vacuum is

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$\\displaystyle{q\\over4\\pi\\varepsilon_0r^2}$

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pointing away radially away from the point charge.

\n

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.

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You might have calculated the force on the second point charge using the full Coulomb force formula,

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$\\displaystyle F={q_1q_2\\over4\\pi\\varepsilon r^2}$ 

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If you approached the question this way you have missed a functional use of the electric field strength: it provides directly the potential for a force to exist, so that $\\vec{F}=q\\vec{E}$.

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Common errors include unit slips and forgetting to square the distance.

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The force due to q1 experienced by q2 (N).

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The second point charge (nC).

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Electric field strength in SI units for a point charge.

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Random distance between 1 and 10 mm in steps of 1 mm (unit are mm)

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Random (point) charge between -10 and +10 nC.

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Permitivitty of free space in F/m.

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Calculate the magnitude of $E$ at a distance {r1} mm from a {q1} nC point-charge.

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$|\\vec{E}|=$ [[0]] V/m.

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A second point charge of magnitude {q2} nC is placed at the distance of {r1} mm from the original {q1} nC point charge.  What is the magnitude of the force it will experience?

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$|\\vec{F}|=$ [[0]] Newtons.

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This question assesses understanding of the names and units of symbols in the Coulomb equation, calculation of quantities and evaluation of force direction.  

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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.

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The Coulomb field is for a point charge (the equation does not generally apply to all charge arrangements):

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
SymbolDescriptionUnits
$E$Electric field strengthV/m (or N/C)
$q$Point chargeC
$\\varepsilon_0$Permittivity of free spaceF/m
$\\varepsilon_r$Relative permittivity of medium betweem the point charge and where the field is being establishednone
$r$Distance from the point charge - it's definitely not a radius!m
$\\hat{r}$This is the unit radial vector.  It is not a displacement (which would have units of m).none
\n

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When we have two point charges, the principle of superposition means that we can determine the electric field strength from each charge separately, and then add these electric fields together, noting that they are vectors and therefore have both magnitude and direciton.  

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The process required for the charges at points A and B, is first to determine at what distance each point charge lies from P.  This value can be substituted in the part of the Coulomb expression that conveys the magnitude (we retain the information in the sign of the charge).  We can then work out the unit vectors from each charge towards P.  For example, the unit vectors pointing from A to P is 

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$\\displaystyle {\\vec{BP}\\over \\left|\\vec{BP}\\right|}$

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Equivalently for the case we have here, the field from B can be expressed using trigonometric functions: 

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$\\displaystyle \\vec{E}_B(P)=\\frac{Q_B}{4\\pi\\varepsilon r_B^2}\\left(\\cos(\\theta)\\hat{i}+\\sin(\\theta)\\right)\\hat{j}, \\tan(\\theta)=\\frac{y(P)-y(B)}{x(P)-x(B)}$.

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Combining the magnitudes and directions for the fields from both point charges gives you the total field at P.

\n

The condition for the point-dipole approximation to be valid is when the distance of the point at which the field is being approximated is far bigger than the distance between two charges of equal magnitude but opposite sign.  This is not the case in this worked example, so the point dipole approximation would be a very poor approach.

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Second point charge in C.

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Permittivity of free space, F/m.

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Point charge in C.

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Charge on the second point charge in muC.

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Point charge in nC.

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Electric field at the distant point in N/C.

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x-co-ordinate of q_A in mm.

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Coulomb's Law is embodied in the equation 

\n

$\\displaystyle \\vec{E}= {q\\over 4 \\pi \\varepsilon_0 \\varepsilon_r r^2}\\hat{r}$.

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$\\vec{E}$ is [[0]] with units [[1]]

\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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"suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

\n

Two point charges are placed in vacuum at the points labelled $A$ and $B$ in the diagram.  The point $P$ is defined, but there is no charge at this point.  The values of the charges, and the positions $x$ and $y$ are shown in the table

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
SiteCharge (nC)$x$ (mm)$y$ (mm)
$A${qan}{xamm}{yamm}
$B${qbn}{xbmm}{ybmm}
$P${xpmm}{ypmm}
\n

Determine the electric field at $P$.

\n

$E_x(P)=$ [[0]] [[2]]

\n

$E_y(P)=$ [[1]] [[2]]

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There is an approximation for the electric field arising from a pair of opposite charges.  Is the point dipole approximation appropriate for the currect case?

\n

[[0]]

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\n

A square of side length {sidecm} cm has charges of different magnitudes placed at its four corners. Clockwise from the top right as viewed from above these are $q_1$ to $q_4$, with values {q1}, {q2}, {q3} and {q4} nC. In this question, you are required to determine the total electric field, including both magnitude and direction, at the centre of the square due to this arrangement of point charges.

\n

This question is a relatively complicated example of application of the principle of superposition.

\n

You should use the values of the physical constants provided in the module notes, assume the medium is air, quote all answers to 4 significant figures and you may express values in scientific format.

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\n

This is a problem of superposition of four different fields, which amounts to the sum of four vectors.  Each field is an expression of Coulomb's Law, so that the pre-factor is

\n

$\\displaystyle {q_i\\over 4\\pi\\varepsilon_0 r^2}$,

\n

where since the point of interest is the centre of the square, the distance $r$ is the same for each point-charge (in this case {side*100}/$\\sqrt{2}\\approx$ {siground(r*100,4)} cm). In each case the field points radially away from the point-charge so that for $q_1$ it is down and left in equal measures, for $q_2$ it is up and to the left, and so on.  For negative charges the sign of $q$ can be thought of reversing the direction (fields converge on negative point charges).  Take care in determining the direction and note that diagrams generally help!

\n

The value of $E$ can be readily obtained and the direction in each case given by

\n

$\\displaystyle{1\\over\\sqrt{2}}\\left(\\pm \\hat{i} \\pm \\hat{j}\\right)$, 

\n

with the combinations of $\\pm$ depending upon which charge we're considering and the $\\sqrt{2}$ factor coming from the fact that this is a unit vector.

\n

Once each vector field has been obtained they can be summed by component (apply the principle of superposition) giving the total $E$-field at the centre of the square.

\n

Finally the angle with the line joining $q_2$ and $q_4$ can be obained by noting that the dot product of the total field with the direction of the displacement from $q_2$ to $q_4$ yields an angle:

\n

$\\displaystyle{\\vec{E}.\\left(-\\frac{\\hat{i}}{\\sqrt{2}}+\\frac{\\hat{j}}{\\sqrt{2}}\\right)=\\left|\\vec{E}\\right|\\cos(\\theta)}$

\n

so

\n

$\\displaystyle{-E_x+E_y=\\sqrt{2}\\left|\\vec{E}\\right|\\cos(\\theta)\\Rightarrow \\theta=\\cos^{-1}\\left(\\frac{E_y-E_x}{\\sqrt{2}\\left|\\vec{E}\\right|}\\right)}$.

\n

It then only remains to determine the value of theta that lies in the required angle range.

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$y$-component of the E-field from q1(N/C).

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Point charge in the bottom left corner, nC.

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Relative permittivity of air.

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Magnitude of the field at the centre.

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Length of side of square in cm.

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$x$-component of the E-field from q2 (N/C).

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Angle between the line between q2 and q4 and the E-field

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$y$-component of the E-field from q4 (N/C).

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$y$-component of the E-field due to q2 (N/C).

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$x$-component of the E-field from q4 (N/C).

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Point charge in the top left corner, nC.

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Distance from the charges to the centre of the square in m.

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Point charge in bottom right corner, nC.

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$x$-component of the E-field from q1 (N/C).

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Permittivity of free space in F/m to 4 significant figures.

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The constant, $1/4\\pi\\varepsilon$ in units of m/F.

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$y$-component of the E-field from q3 (N/C).

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$x$-component of the E-field from q3 (N/C).

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Point charge in top right corner in nC.

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r^2, cm

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Side length in m.

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q1 in C

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q2 in C

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q3 in C

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q4 in C

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Determine the electric field at the centre of the square due to $q_1$ ({q1}nC).

\n

$E_x(q_1)=$[[0]]V/m

\n

$E_y(q_1)=$[[1]]V/m

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Determine the electric field at the centre of the square due to $q_2$ ({q2}nC).

\n

$E_x(q_2)=$[[0]]V/m

\n

$E_y(q_2)=$[[1]]V/m

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Determine the electric field at the centre of the square due to $q_3$ ({q3}nC).

\n

$E_x(q_3)=$[[0]]V/m

\n

$E_y(q_3)=$[[1]]V/m

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Determine the electric field at the centre of the square due to $q_4$ ({q4}nC).

\n

$E_x(q_4)=$[[0]]V/m

\n

$E_y(q_4)=$[[1]]V/m

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Determine the total electric field at the centre of the square due to the all four charges.

\n

$E_x=$[[0]]V/m

\n

$E_y=$[[1]]V/m

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What is the angle, $\\theta$, between the direction of the $E$-field and the line joining the charges $q_2$ ({q2}nC) and $q_4$ ({q4}nC)?  Express your answer as an angle in degrees in the range $0^\\circ\\le\\theta\\le90^\\circ$.

\n

$\\theta=$[[0]]$^\\circ$.

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The electrostatic potential due to a point charge is calculated at three points, who of which are at the same distance but different directions.  This relates to the idea that the equipoptentials of a point charge are spheres centred on the charge, so all points at the same distance are at the same potential.

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. $V(\\infty)=0\\,\\text{V}$.

\n

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.

", "advice": "

Quantiative errors may occur due to unit conversion errors.

\n

A 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.

\n

The correct formula for the electrostatic potential due to a point charge, taking the boudary condition that $V(\\infty)=0$, is 

\n

$\\displaystyle V(r)=\\frac{q}{4\\pi\\varepsilon r}$

\n

where $r$ is the distance from the point charge.  Note, the electrostatic potential is a scalar field - it does not have a direction.

\n

If 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.

\n

The 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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$1/4\\pi\\varepsilon_0$ and a factor of 10,000 to convert cm and microC to get V in SI units.

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Electrostatic potential at (x1,x1) in V.

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Randomised value of the charge at the origin in $mu$C.

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Electrostatic potential at (x1,x2) in V.

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Distance of point c from the charge in cm.

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First $x$-location in cm.

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Electrostatic potential at (x2,x1) in V.

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Permittivity of free space (F/m) to four significant figures.

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Distance of point b from the charge in cm.

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Second $x$-position in cm.

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Distance of point a from the charge in cm.

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Primitive charge

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What is the electrostitic potential at the point $a$, with co-ordinates $x=${x1} cm, $y=${x2} cm, $z=$0 cm for a point charge of {q1} $\\mu$C located at the origin?

\n

$r_a=$ [[1]] cm.

\n

$V(r_a)=$[[0]] Volts.

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What is the electrostitic potential at the point $b$, with co-ordinates $x=${x1} cm, $y=${x1} cm, $z=$0 cm for a point charge of {q1} $\\mu$C located at the origin?

\n

$r_b=$ [[1]] cm.

\n

$V(r_b)=$[[0]] Volts.

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What is the electrostitic potential at the point $c$, with co-ordinates $x=${x2} cm, $y=${x1} cm, $z=$0 cm for the point charge of {q1} $\\mu$C located at the origin?

\n

$r_c=$ [[1]] cm.

\n

$V(r_c)=$[[0]] Volts.

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Two of the three values of $V(r_a)$, $V(r_b)$ and $V(r_c)$ have the same value.  This is because...

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A point charge of {q2} μC point is initially placed at $c$.  The repulsive force moves it far from the origin.  Taking it's final position to be effectively infinitly far from the {q1} μC point charge at the origin, how much energy does this second charge gain in this process?

\n

Energy gained = [[0]] J

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Gauss' Law is a powerful, general expression that links the arrangement of charges to the electric field strength, $\\vec{E}$.  The mathematical formulation of the Law links the electric field flux through a closed surface, $\\phi_{\\rm electric}$ to the charge contained within that surface, $q$.

\n

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.

", "advice": "

This is partly a mathematical and partly a computational problem, but it centres upon the formulation of Gauss' Law as 

\n

$\\displaystyle \\phi_{\\rm electric}=\\oint_S \\vec{E}.d\\vec{A}={q\\over\\varepsilon}$,

\n

where $\\phi_{\\rm electric}$ is the electric field flux through the closed Gaussian surface, $S$, $\\vec{E}$ is the electric field through $S$, $d\\vec{A}$ is an element of the surface area, $q$ is the total charge containined within $S$ and $\\varepsilon$ is the permittivity.  

\n

To address the numerical part, we simply note that the arrangement of the charge is not relevant (Gauss' law applies whatever the arrangement of the charge within the closed surface) and so we can use  

\n

$\\displaystyle |\\phi_{\\rm electric}|=\\left|{q\\over\\varepsilon}\\right|=\\left|{e N_{\\rm electron}\\over\\varepsilon}\\right|$,

\n

where $e=1.6\\times10^{-19}$C is the primitive charge.

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flux in V.m

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Number of electrons within the Gaussian surface.

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Permittivity of free space, F/m

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Charge on an electron, C.

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Number of electrons inside the Gaussian surface.

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Charge contained within the Gaussian surface in C.

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Flux through Gaussian surface in V.nm

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The flux through the surface expressed as an integral in terms of the electric field strength $\\vec{E}$ (NUMBAS input Efield) and the elements of the surface, $d\\vec{A}$ (NUMBAS input dA) is an integral over the closed surface, $S$, expressed as 

\n

$\\displaystyle \\phi=\\oint_S$[[0]]

\n

Your answer should be input as either a dot prodct using 'dot(a,b)' where a and b are the symbols for the vectors, or a cross product using 'cross(a,b)'.

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The flux through the surface is equal to a quantity depending upon the charge contained within the surface, $q$, and the permittivity of the space containing the charge, $\\varepsilon$ (NUMBAS input epsilon).  This quantity is

\n

[[0]]

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When selecting a Gaussian surface one might take a number of factors into account.  Indicate at least 1 typical condition.  Note well - marks are deducted for inappropriate answers.

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The electric field from an arrangement of charge is found to give rise to an electric flux through a Gaussian surface of {fluxnm} volt-nanometres.   If the charge is otherwise in vacuum, what is the total amount of charge inside the Gaussian surface?

\n

$|Q|=$ [[0]] Coulombs

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If the charge in the previous question is due to a collection of electrons, how many are inside the Gaussian surface?

\n

$n_{\\rm electrons}=$ [[0]]

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Estimate the electric field strength, $E$, required to make a balloon stick to a ceiling due to electrostatic force. Assume that the balloon has a mass of {massg} g and has an area of contact with the ceiling of {areacm} cm$^2$, and note that the force due to gravity is $mg$ (pointing towards the centre of the Earth).

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.

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The force due to gravity, or weight, at the Earth's surface is written as $mg$, where $m$ is the mass and $g$ is the acceleration due to gravity.  The electric field force can be either written in terms of the force on a charge ($F=qE$), or, as is required in this case, in terms of the energy stored in the field, $F=\\displaystyle{\\varepsilon E^2 A\\over 2}$, where $A$ is the contact area, $\\varepsilon$ is the permittivity of air (a balloon stuck to the ceiling is likely to be in air!) and $E$ is the electric field strength.  For NUMBAS we need to express the electric field as a different symbol as $E$ is reserved as {e}, the base of the natural logarithm.  

\n

For the electric field to just balance the weight we can equate the two forces and re-arrange for $E$:

\n

$\\displaystyle{\\varepsilon E^2 A\\over 2}=mg\\Rightarrow E=\\sqrt{\\displaystyle{2mg\\over \\varepsilon A}}$.

\n

You have to remember that the equations work for SI units, so the mass should be in kg and the area in m$^2$.

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Area in m^2

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Relative permittivity of air.

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Mass of the balloon in g.

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Acceleration due to gravity on Earth, m/s^2

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Computed E-field that JUST balances the weight.  

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Mass in kg.

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Permittivity of free space, F/m.

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Area of contact in cm^2

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The force due to gravity, i.e. the weight of the balloon is given by the expression

\n

$F_{\\rm gravity}=$ [[0]]

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The force from the elecrtic field opposing the weight of the balloon is given by an expression in terms of the permiitivity, $\\varepsilon$ (epsilon), the contact area, $A$ and the electric field strength, $E_1$ (E_1):

\n

$F_{\\rm electrostatic}=\\frac{\\varepsilon E_1^2 A}{2}$ 

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Calculate the value of $E$ that balances the weight.

\n

$E=$[[0]] V/m

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There are unattempted question parts.  If this is not intended you should review the questions to find the part or parts that have not been successfully submitted.

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