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Elecrostatics flash cards

Master Elecrostatics through 90 JEE Advanced-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Elecrostatics, question and answer

15 of this chapter's 90 cards, laid out open so you can read straight through. The remaining 75 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.State Coulomb's law in vector form for the force on charge q1q_1 due to q2q_2.

    F12=14πε0q1q2r2r^21\vec{F}_{12}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}\,\hat{r}_{21}, where r^21\hat{r}_{21} points from q2q_2 to q1q_1. Like charges repel, unlike attract.

    Hint: Inverse-square, along the line joining the charges.

  2. 2.What is the value of 14πε0\dfrac{1}{4\pi\varepsilon_0} in SI units?

    14πε09×109 N m2/C2\dfrac{1}{4\pi\varepsilon_0}\approx 9\times10^{9}\ \text{N m}^2/\text{C}^2, with ε0=8.85×1012 C2/N m2\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/\text{N m}^2.

    Hint: Coulomb constant kk.

  3. 3.How does the electrostatic force between two point charges change if a dielectric of constant KK fills the space?

    The force reduces by a factor KK: Fmedium=FvacuumK=14πε0Kq1q2r2F_{medium}=\dfrac{F_{vacuum}}{K}=\dfrac{1}{4\pi\varepsilon_0 K}\dfrac{q_1q_2}{r^2}.

    Hint: Replace ε0\varepsilon_0 with ε=Kε0\varepsilon=K\varepsilon_0.

  4. 4.State the principle of superposition for electrostatic forces.

    The net force on a charge is the vector sum of the individual Coulomb forces from every other charge, each computed as if the others were absent.

    Hint: Forces add as vectors, independently.

  5. 5.Define the electric field E\vec{E} at a point.

    E=limq00Fq0\vec{E}=\displaystyle\lim_{q_0\to0}\dfrac{\vec{F}}{q_0} — force per unit positive test charge. Units: N/C\text{N/C} or V/m\text{V/m}.

    Hint: Test charge taken small so it doesn't disturb the source.

  6. 6.Electric field due to a point charge qq at distance rr?

    E=14πε0qr2r^\vec{E}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}\,\hat{r}. Points radially outward for q>0q>0, inward for q<0q<0.

    Hint: Same inverse-square shape as Coulomb force.

  7. 7.List key properties of electric field lines.

    Start on ++ charges, end on -; never cross; tangent gives E\vec{E} direction; density \propto field magnitude; no closed loops in electrostatics; perpendicular to conductor surfaces.

    Hint: Density and direction encode the field.

  8. 8.Why can two electric field lines never intersect?

    At an intersection the field would have two directions at once, which is impossible since E\vec{E} is uniquely defined at each point.

    Hint: Tangent = field direction, must be single-valued.

  9. 9.Define electric dipole moment p\vec{p}.

    p=qd\vec{p}=q\,\vec{d} (magnitude q×2aq\times 2a for separation 2a2a), directed from the negative to the positive charge. Units: Cm\text{C\,m}.

    Hint: Points from q-q to +q+q.

  10. 10.Field on the axial line of a dipole at distance rr (rar\gg a)?

    Eaxial=14πε02pr3E_{axial}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}, directed parallel to p\vec{p}.

    Hint: Twice the equatorial value, 1/r31/r^3.

  11. 11.Field on the equatorial line of a dipole at distance rr (rar\gg a)?

    Eeq=14πε0pr3E_{eq}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}, directed antiparallel to p\vec{p}.

    Hint: Half the axial value, opposite direction.

  12. 12.How does a dipole's field fall off with distance compared to a point charge?

    Dipole field 1/r3\propto 1/r^3, while a point charge field 1/r2\propto 1/r^2. The dipole falls faster.

    Hint: Net charge zero speeds the decay.

  13. 13.Torque on a dipole p\vec{p} in a uniform field E\vec{E}?

    τ=p×E\vec{\tau}=\vec{p}\times\vec{E}, magnitude τ=pEsinθ\tau=pE\sin\theta. It tends to align p\vec{p} with E\vec{E}.

    Hint: Cross product; zero when aligned or anti-aligned.

  14. 14.Net force on a dipole in a uniform field?

    Zero — the forces +qE+q\vec{E} and qE-q\vec{E} are equal and opposite. Only a torque acts (unless the field is non-uniform).

    Hint: Equal & opposite forces on the two ends.

  15. 15.Potential energy of a dipole in a uniform field?

    U=pE=pEcosθU=-\vec{p}\cdot\vec{E}=-pE\cos\theta. Minimum (pE-pE) when aligned, maximum (+pE+pE) when anti-aligned.

    Hint: Zero at θ=90\theta=90^\circ by convention.

Open the interactive deck for the other 75 cards, with self-grading so the ones you keep missing come back.

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