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McGill University · PHYS 142

McGill PHYS 142: Electromagnetism and Optics tutoring

Work through electromagnetism and optics questions in PHYS 142 with an online tutor. For electrostatics, your tutor can help you describe the charge distribution, draw field directions and decide which method fits before calculating. You can start with a question or a concept.

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Online 60 min: $55 CAD · PricingMidterm & final exam helpTry Gauss’s law if it is in your outlineWarm-up: field at the midpointFree lesson: electric field and potential
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Course overview

McGill PHYS 142 covers basic electricity and magnetism together with geometrical optics. It forms part of McGill’s calculus-based introductory physics sequence.

PHYS 142 topics to review

Electric charge and fieldsElectric potentialCircuitsMagnetic fieldsElectromagnetic inductionGeometrical optics

Common difficulties

  • Using vector addition for fields and scalar addition for potential
  • Justifying a symmetry shortcut before applying it

Preparing for this course

Combine vectors, mechanics, trigonometry and elementary calculus. Review how a field differs from the force on a particular charge.

Registration requirements

The guidance above is study preparation, not an eligibility decision. Confirm current prerequisites, restrictions and availability with McGill University.

Check the official course catalogue

Try a course problem

Two identical positive point charges lie equally far left and right of their midpoint. What is their net electric field at that midpoint?

Show the reasoning

Each positive charge produces a field directed away from itself. At the midpoint the left charge’s field points right and the right charge’s field points left. Equal charge and equal distance give equal magnitudes, so the vector sum is zero. Their scalar potentials, referenced to zero at infinity, are both positive and add rather than cancel.

Which charge is inside your Gaussian surface?

Use this example when Gauss’s law is in your current PHYS 142 outline. This is original topic practice, not an official assessment question.

An isolated, nonconducting solid sphere of radius R=0.20 mR=0.20\,\mathrm{m} carries total charge Q=+8.0 nCQ=+8.0\,\mathrm{nC} spread uniformly throughout its volume. The charge stays fixed. Neglect polarization, use vacuum permittivity throughout, and take k=14πε0=8.99×109 N m2/C2.\begin{aligned}k&=\frac{1}{4\pi\varepsilon_0}\\&=8.99\times10^9\,\mathrm{N\,m^2/C^2}.\end{aligned} Find the electric field at distances r=0.10 mr=0.10\,\mathrm{m} and r=0.40 mr=0.40\,\mathrm{m} from the centre. State both magnitude and direction.

A solid insulating sphere has uniformly distributed volume charge of +8.0 nC and radius R = 0.20 m. A dashed concentric Gaussian sphere of radius r = 0.10 m lies inside the solid boundary. Only the charged volume inside that Gaussian surface contributes to the enclosed charge.
Interior: enclosed volume only.
The same solid insulating sphere has radius R = 0.20 m and uniform volume charge +8.0 nC. A dashed concentric Gaussian sphere of radius r = 0.40 m surrounds the solid boundary, enclosing all +8.0 nC.
Exterior: the Gaussian sphere encloses all the volume charge.

Predict first: The nearer point is inside the charged material. Does a sphere of radius 0.10 m enclose all the charge? Should the field become infinite at the centre?

Show a hint

The physical sphere has fixed radius R. Your Gaussian surface has variable radius r. Uniform volume charge makes the enclosed fraction equal to the enclosed volume fraction.

Show the reasoning

  1. Establish the symmetry. The charge density depends only on distance from the centre; there is no external field. Rotations leave the distribution unchanged. The electric field must therefore be radial and have the same magnitude everywhere on a concentric sphere. For this positive charge it points outward. This is the justification for simplifying the flux integral, not a consequence of drawing a spherical surface by itself.

  2. Write Gauss’s law with the correct enclosed charge. The outward flux through a concentric sphere is E(r)4πr2=Qencε0E(r)4\pi r^2=\frac{Q_{\mathrm{enc}}}{\varepsilon_0}. For 0<r<R0<r<R, uniform density gives Qenc=Q(rR)3Q_{\mathrm{enc}}=Q\left(\frac{r}{R}\right)^3. For r≥Rr\ge R, Qenc=QQ_{\mathrm{enc}}=Q.

  3. Solve in the two regions. Inside, E(r)=kQrR3E(r)=\frac{kQr}{R^3}; outside, E(r)=kQr2E(r)=\frac{kQ}{r^2}. At the centre E=0E=0 by symmetry and by the limit of the inside formula. At r=Rr=R, both expressions give kQR2\frac{kQ}{R^2}, so the field is continuous for this volume-charge model.

  4. Evaluate the interior point. At r=0.10 mr=0.10\,\mathrm{m}, the Gaussian sphere encloses (12)3=18\left(\frac12\right)^3=\frac18 of the charge: Qenc=+1.0 nCQ_{\mathrm{enc}}=+1.0\,\mathrm{nC}. Thus E=k(1.0×10−9)(0.10)2=899 N/C,\begin{aligned}E&=\frac{k(1.0\times10^{-9})}{(0.10)^2}\\&=899\,\mathrm{N/C},\end{aligned} radially outward. Using all 8.0 nC here would overestimate the field by a factor of eight.

  5. Evaluate the exterior point. At r=0.40 mr=0.40\,\mathrm{m}, all the charge is enclosed. E=k(8.0×10−9)(0.40)2=449.5 N/C,\begin{aligned}E&=\frac{k(8.0\times10^{-9})}{(0.40)^2}\\&=449.5\,\mathrm{N/C},\end{aligned} radially outward. To two significant figures the results are 9.0×102 N/C9.0\times10^2\,\mathrm{N/C} and 4.5×102 N/C4.5\times10^2\,\mathrm{N/C}.

Electric field magnitude of the uniformly volume-charged solid sphere versus radius. It rises linearly from zero at the centre to the continuous peak at radius R with field kQ/R squared, then falls as one over r squared. The normalized peak is at r/R = 1 and E/(kQ/R squared) = 1.
For the original solid sphere, the two expressions meet at (R,kQR2)\left(R,\frac{kQ}{R^2}\right).

Common mistake: Qenc=0Q_{\mathrm{enc}}=0 implies zero total flux. It does not, by itself, imply zero electric field at every point on the surface. You also need the symmetry argument. For example, a uniform nonzero external field has zero net flux through a closed sphere because as much field enters as leaves.

Your turn 1: check the boundary

Find the field at r=Rr=R and compare the interior and exterior results before substituting values.

Show the explained answer

Both give E(R)=kQR2=1798 N/C,\begin{aligned}E(R)&=\frac{kQ}{R^2}\\&=1798\,\mathrm{N/C},\end{aligned} about 1.8×103 N/C1.8\times10^3\,\mathrm{N/C} outward. The field rises linearly from the centre to the surface, then falls as 1/r21/r^2. This comparison tests whether the two formulas describe the same distribution.

Your turn 2: change the distribution

Keep Q=+8.0 nCQ=+8.0\,\mathrm{nC} and R=0.20 mR=0.20\,\mathrm{m}, but put the charge uniformly on an ideal infinitesimally thin spherical shell. Find the fields at r=0.10 mr=0.10\,\mathrm{m} and r=0.40 mr=0.40\,\mathrm{m}. Do not evaluate exactly on the charged surface.

An ideal thin spherical shell of radius R = 0.20 m carries all +8.0 nC uniformly on its surface. A dashed concentric Gaussian sphere of radius r = 0.10 m lies inside the shell. Compare this surface-charge distribution with the original volume-charged solid sphere.
Changed distribution: the shell carries charge only on its surface.
Show the explained answer

At the interior point, a concentric Gaussian sphere encloses no charge. Spherical symmetry still makes E constant in magnitude and radial, so Gauss’s law gives E=0E=0. Outside, the field remains 449.5 N/C outward because the total enclosed charge is unchanged. The interior field differs because the distribution differs. A uniformly volume-charged insulator is not interchangeable with a shell or a conducting sphere in electrostatic equilibrium.

Your turn 3: separate field from force

Return to the original solid sphere. Put a test charge q=−2.0 nCq=-2.0\,\mathrm{nC} at r=0.10 mr=0.10\,\mathrm{m}. Neglect any disturbance of the source charge. Find the radial component of its force, taking outward as positive.

Show the explained answer

Fr=qErF_r=qE_r, with Er=+899 N/CE_r=+899\,\mathrm{N/C}. Thus Fr=(−2.0×10−9)(+899)=−1.798×10−6 N,\begin{aligned}F_r&=(-2.0\times10^{-9})(+899)\\&=-1.798\times10^{-6}\,\mathrm{N},\end{aligned} about −1.8 μN. The negative sign means inward. The electric field remains outward; changing the test charge changes the force on that charge, not the source field.

How tutoring can help

You can start with the electrostatics question, even if you have not chosen a surface or written an equation. Your tutor can help you describe the charge distribution, draw the field directions and check whether symmetry makes Gauss’s law a useful method. You can then choose a surface, find the enclosed charge and check the result together. If you need an earlier starting point, your tutor can explain the difference between electric field, potential and force using the midpoint example.

Exam preparation for PHYS 142

Start each electrostatics question by naming the required quantity and the charge distribution. For a Gauss’s-law question on your review list, justify the symmetry before replacing a flux integral by a product, and check the enclosed charge separately in each region. Retry the shell and test-charge questions without the solutions. Use your section’s assigned material to add circuit, magnetic and geometrical-optics questions; these exercises cover only selected electrostatics skills.

Use your section’s outline to choose the chapters to review. We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

Useful links for preparation or further study—not transfer equivalencies or a required registration sequence.

PHYS 131 tutoringMechanics and WavesMATH 141 tutoringCalculus 2PHYS 102 tutoringIntroductory Physics — Electromagnetism
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PHYS 142 · official course reference