Skip to main content
NEET Test Series — Practice smart, score high.

Elecrostatics flash cards

Master Elecrostatics through 100 NEET-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

22 of this chapter's 100 cards, laid out open so you can read straight through. The remaining 78 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.State Coulomb's law for two point charges in vacuum.

    The force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them: F=14πε0q1q2r2F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}. It acts along the line joining the charges.

    Hint: Product of charges over r2r^2.

  2. 2.What is the value of the Coulomb constant k=14πε0k=\dfrac{1}{4\pi\varepsilon_0} in SI units?

    k9×109 N m2C2k \approx 9\times 10^{9}\ \text{N m}^2\,\text{C}^{-2}. Here ε0=8.85×1012 C2N1m2\varepsilon_0 = 8.85\times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2} is the permittivity of free space.

    Hint: 9×1099\times10^9.

  3. 3.How does Coulomb's law change when charges are placed in a medium of dielectric constant KK?

    The force is reduced by a factor KK: Fmed=14πε0Kq1q2r2=FvacKF_{med}=\dfrac{1}{4\pi\varepsilon_0 K}\dfrac{q_1 q_2}{r^2}=\dfrac{F_{vac}}{K}. The medium's permittivity is ε=Kε0\varepsilon = K\varepsilon_0.

    Hint: Divide vacuum force by KK.

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

    The net force on a charge due to several charges is the vector sum of the forces exerted by each charge individually, each computed as if the others were absent. Forces add independently (they do not screen one another).

    Hint: Vector sum of pairwise forces.

  5. 5.What is meant by quantisation of charge?

    Charge exists only in integer multiples of the elementary charge: q=neq=ne, where nn is an integer and e=1.6×1019 Ce=1.6\times10^{-19}\ \text{C}. Charge cannot be a fraction of ee in observable bodies.

    Hint: q=neq=ne.

  6. 6.State the law of conservation of charge.

    The total charge of an isolated system remains constant. Charge can be transferred from one body to another but can neither be created nor destroyed.

    Hint: Net charge of isolated system is fixed.

  7. 7.What is additivity of charge?

    Total charge of a body is the algebraic (scalar) sum of all the individual charges on it, taking signs into account. Charge is a scalar quantity.

    Hint: Algebraic sum, with sign.

  8. 8.Define electric field intensity E\vec{E} at a point.

    It is the force per unit positive test charge placed at that point: E=limq00Fq0\vec{E}=\lim_{q_0\to 0}\dfrac{\vec{F}}{q_0}. SI unit: N/C\text{N/C} or V/m\text{V/m}. It is a vector directed along the force on a positive charge.

    Hint: Force per unit positive charge.

  9. 9.Write the 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}. Its magnitude is E=kqr2E=\dfrac{kq}{r^2}, directed radially outward for positive qq and inward for negative qq.

    Hint: kq/r2kq/r^2 radial.

  10. 10.Why is the test charge taken to be vanishingly small when defining E\vec{E}?

    A finite test charge would exert its own force on the source charges and disturb their configuration, altering the very field being measured. The limit q00q_0\to 0 avoids this disturbance.

    Hint: Avoid disturbing the source distribution.

  11. 11.List the key properties of electric field lines.

    They start on positive charges and end on negative charges; the tangent gives the field direction; they never intersect; density (closeness) indicates field strength; they do not form closed loops in electrostatics; they are perpendicular to conductor surfaces.

    Hint: Start +, end −, never cross.

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

    At the point of intersection the field would have two directions (two tangents), which is impossible since E\vec{E} has a unique direction at every point.

    Hint: Field has a single direction per point.

  13. 13.What does the relative closeness of field lines indicate?

    Where field lines are crowded, the field is strong; where they are widely spaced, the field is weak. The number of lines per unit area (normal) is proportional to E|\vec{E}|.

    Hint: Crowded lines = strong field.

  14. 14.Why are electrostatic field lines always perpendicular to a conductor's surface?

    If there were a tangential component of E\vec{E} at the surface, free charges would move along it until equilibrium. In electrostatics no charge moves, so E\vec{E} must have no tangential component, i.e. it is normal to the surface.

    Hint: No tangential field in equilibrium.

  15. 15.Why do electrostatic field lines never form closed loops?

    The electrostatic field is conservative, so the work done around any closed path is zero. A closed field line would imply nonzero work moving a charge around it, contradicting conservativeness.

    Hint: Conservative field.

  16. 16.Define an electric dipole and its dipole moment.

    An electric dipole is a pair of equal and opposite charges +q+q and q-q separated by a small distance 2a2a. Its dipole moment is p=q(2a)\vec{p}=q(2a), directed from the negative to the positive charge. Unit: C m\text{C m}.

    Hint: p=q×2ap=q\times 2a, points −q to +q.

  17. 17.Write the electric field on the axial line of a short dipole.

    Eaxial=14πε02pr3E_{axial}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2p}{r^3}, directed along p\vec{p} (parallel to the dipole moment). Valid for rar\gg a.

    Hint: 2kp/r32kp/r^3, along pp.

  18. 18.Write the electric field on the equatorial line of a short dipole.

    Eeq=14πε0pr3E_{eq}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^3}, directed antiparallel to p\vec{p} (opposite to the dipole moment). Valid for rar\gg a.

    Hint: kp/r3kp/r^3, opposite to pp.

  19. 19.What is the ratio of axial to equatorial field of a short dipole at the same distance?

    EaxialEeq=2\dfrac{E_{axial}}{E_{eq}}=2. The axial field is twice the equatorial field at equal distances, and the two are oppositely directed relative to p\vec p.

    Hint: Factor of 2.

  20. 20.Why does a dipole's field fall off as 1/r31/r^3 rather than 1/r21/r^2?

    The fields of +q+q and q-q nearly cancel at large distances; the residual is a small difference that decreases faster, giving a net 1/r31/r^3 dependence for a neutral dipole.

    Hint: Partial cancellation of opposite charges.

  21. 21.Write the torque on a dipole in a uniform electric field.

    τ=p×E\vec{\tau}=\vec{p}\times\vec{E}, with magnitude τ=pEsinθ\tau=pE\sin\theta, where θ\theta is the angle between p\vec{p} and E\vec{E}. It tends to align p\vec{p} with E\vec{E}.

    Hint: τ=pEsinθ\tau=pE\sin\theta.

  22. 22.What is the net force on a dipole placed in a uniform electric field?

    Zero. The forces +qE+q\vec{E} and qE-q\vec{E} are equal and opposite, so they cancel — but they form a couple giving a net torque (unless θ=0\theta=0 or π\pi).

    Hint: Forces cancel; torque may not.

Open the interactive deck for the other 78 cards, with self-grading so the ones you keep missing come back.

Other ways to revise this chapter

Master this chapter with similar other learning materials.

Preparing students for India’s top institutes

Our students are currently into top technological and medical institutes of India.

  • IIT Bombay
  • IIT Delhi
  • IIT Madras
  • IIT Kanpur
  • IIT Kharagpur
  • IIT Roorkee
  • IIT Guwahati
  • IIT BHU Varanasi
  • AIIMS Delhi
  • NIT Tiruchirappalli
  • NIT Rourkela

Join QuestPix, Today!

Get notified first, with exam & curriculum updates, course & test series launch offers, motivation & success stories and free learning resources recommended by toppers.

Chat on WhatsApp