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Electric Charges and Fields

Electric Charges and FieldsNEET Physics · Class 12 · NCERT Chapter 1

7 interactive concept widgets for Electric Charges and Fields. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

Coulomb force and electric field

Force between two charges, field of a single point charge, and vector superposition of two charges.

Coulomb's law

Coulomb's law calculator

Force between two point charges scales with the product of charges and inversely with the square of distance.

Force between two point charges. Positive charges in microcoulombs (μC); negative values give attraction.

q₁: 2.00 μC

q₂: 3.00 μC

r: 10.0 cm

Force F (magnitude)

5.400 N

Repulsive (same sign)

Try this

  • Halve r: force becomes 4 times larger (1/r² scaling).
  • Set q₁ and q₂ with opposite signs: force becomes attractive (green).
  • 1 C is enormous: even 1 μC at 1 cm gives a noticeable force on a similar charge.
Electric field

Electric field of a point charge

The field at distance r from a point charge q is k|q| over r squared. Direction is set by the sign of q.

Field from a point charge falls off as 1/r². Direction is radially outward for positive charge, radially inward for negative charge.

Charge q: 5.00 μC

Distance r: 50.0 cm

Electric field magnitude

1.800e+5 V/m

Direction: radially outward from the charge

Try this

  • Doubling r reduces E by a factor of 4.
  • Field of 1 μC at 1 cm: about 9 x 10⁷ V/m. Massive.
  • Negative q: same magnitude formula but field points TOWARD the charge.
Superposition

Two-charge superposition

Total field at any point is the vector sum of the fields from each charge. Slide the test point and watch the resultant arrow.

Two charges fixed at x = -0.5 m and x = +0.5 m. Move the test point and watch the resultant field vector.

q₁ (left): 2.00 μC

q₂ (right): -2.00 μC

Test point x: 0.50 m

Test point y: 0.30 m

Net field at point

|E| = 1.96e+5 V/m

E_x = 1.58e+4 V/m, E_y = -1.95e+5 V/m

+

Try this

  • Equal and opposite charges (+q and -q): on the perpendicular bisector, E points from + to -.
  • Equal positive charges: on the perpendicular bisector, fields combine to point straight away.
  • Set the test point right between two equal positive charges: net E = 0.

Flux and Gauss's law

Enclosed-charge view of flux, plus three classic Gauss applications: line, plane and shell.

Gauss's law

Electric flux and Gauss's law

Total flux through a closed surface depends only on the enclosed charge. The shape of the surface does not matter.

Total electric flux through any closed surface depends only on the charge ENCLOSED, not on the shape or size of the surface.

Enclosed charge q: 5.00 μC

Total flux through closed surface

5.647e+5 V·m

Try this

  • Place the charge inside any shape (cube, sphere, blob), flux is the same Q over epsilon_0.
  • Charge OUTSIDE the surface: net flux through it is zero (entering and leaving cancel).
  • Negative charge enclosed: net flux is negative (field lines come in).
  • 1 μC enclosed: flux about 1.13 x 10⁵ V·m.
Gauss applications

Field via Gauss's law (line, plane, shell)

Three NEET-favourite geometries solved by symmetry plus Gauss's law.

Three classic Gauss's law geometries plus the inside of a charged shell. Pick a shape, set the source density and distance.

Linear charge density λ (μC/m): 5.00

Distance from line r: 0.50 m

Electric field magnitude

1.798e+5 V/m

Try this

  • Infinite line: E falls as 1/r (slower than a point charge).
  • Infinite plane: E does NOT depend on distance. Field is uniform.
  • Outside a shell: looks just like a point charge at the centre. Like Newton's shell theorem.
  • Inside a shell: E = 0 everywhere. The shell provides full electrostatic shielding.

Electric dipole

Field on the axis vs equator of a dipole, and the torque it feels in a uniform field.

Dipole field

Electric dipole field

Field on the axis vs the equator of a dipole. Two NEET-favourite results, side by side.

Field of an electric dipole at far distances (r much greater than dipole length 2 a). Two famous results: axial is twice as strong as equatorial, and both fall off as 1/r³.

Dipole moment p: 1.00 nC·m

Distance r: 0.50 m

Field at distance r

1.440e+2 V/m

Try this

  • At the same r, axial field is exactly twice the equatorial field.
  • Both fall as 1/r³, faster than a point charge's 1/r².
  • Direction: axial field is along p (from -q to +q). Equatorial is opposite to p.
Dipole torque

Torque and PE of a dipole in a uniform field

A dipole in an external field feels a torque trying to align p with E. The associated potential energy is -p·E.

Torque on a dipole in a uniform field tries to align p with E. Maximum torque at theta = 90°, zero at 0° and 180°.

Dipole moment p: 2.0 nC·m

Field E: 1.0 × 10⁵ V/m

Angle θ between p and E: 45°

Torque τ

1.41e-4 N·m

Potential energy U

-1.41e-4 J

E →+p

Try this

  • θ = 0° (p parallel to E): torque = 0, U is minimum (stable equilibrium).
  • θ = 90° (perpendicular): maximum torque, U = 0.
  • θ = 180° (anti-parallel): torque = 0, U is maximum (UNSTABLE equilibrium, tiniest nudge will flip it).
  • Work to rotate from θ₁ to θ₂: W = pE(cos θ₁ − cos θ₂).

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