Electric Charges and Fields

1. Electric Charge

Definition: Intrinsic property of elementary particles of matter which gives rise to electric force between various objects.

Properties of Charge

Quantization: Q = ne, where n is an integer and e = 1.6 × 10^-19 C.
Conservation: Total charge of an isolated system remains constant.
Additivity: Total charge is the algebraic sum of individual charges (Scalar).
Invariance: Magnitude of charge does not depend on the speed of the particle (unlike mass).

2. Coulomb's Law

Force between two point charges is directly proportional to the product of magnitude of charges and inversely proportional to the square of the distance between them.

F = k |q₁q₂| / r²

k = 1 / 4πε₀ = 9 × 10⁹ N m² C^-2

Dielectric Constant (K or ε_r)

When charges are placed in a medium, force decreases.

F_medium = F_air / K

(For metals, K = ∞, so force becomes zero).

(Asked in NEET 2016)

3. Electric Field

Force experienced per unit positive test charge.

E = F / q₀
Due to point charge: E = kQ / r²

Properties of Field Lines

Start from positive charge, end at negative charge.
Tangent gives direction of Electric Field.
Two lines never cross each other (otherwise two directions at one point).
Electrostatic field lines do not form closed loops (Conservative nature).

4. Electric Dipole

System of two equal and opposites charges separated by a small distance 2a.

Dipole Moment (Vector): p = q × 2a
Direction: Negative to Positive.

Position	Electric Field (short dipole)	Direction
Axial	2kp / r³	Along p
Equatorial	kp / r³	Opposite to p

Note: E_axial = 2 E_equatorial

Dipole in Uniform Field

Net Force F_net = 0
Torque τ = pE sinθ (Vector: τ = p × E)
Max Torque at θ = 90°.

(Asked in NEET 2013, 2020)

5. Electric Flux & Gauss's Law

Electric Flux (Φ)

Total number of field lines crossing an area normally.

Φ = E ⋅ A = EA cosθ
(θ is angle between Field and Area Vector).

Gauss's Law

Total flux through a closed surface is 1/ε₀ times the net enclosed charge.

Φ_closed = ∮ E ⋅ dA = q_in / ε₀

Applications

Infinitely Long Wire: E = λ / 2πε₀r (E ∝ 1/r)
Infinite Plane Sheet: E = σ / 2ε₀ (Independent of distance r)
Thin Spherical Shell:
- Outside (r > R): E = kQ / r²
- Inside (r < R): E = 0

(Asked in NEET 2018, 2019)

Numericals: Electric Charges

Q1. A polythene piece rubbed with wool is found to have a negative charge of 3 × 10^-7 C. Estimate the number of electrons transferred.

Solution:
Q = ne
n = Q / e
n = (3 × 10^-7) / (1.6 × 10^-19)
n ≈ 1.875 × 10¹² electrons.
(Transferred from wool to polythene).

Q2. Two point charges q_A = 3 μC and q_B = -3 μC are located 20 cm apart in vacuum. Find electric field at the midpoint O.

Solution:
Distance r = 10 cm = 0.1 m.
Total field at O = E_A + E_B (Both point towards B because A is +ve repelling and B is -ve attracting).
E = kQ/r²
E_net = 2 × (9 × 10⁹) × (3 × 10^-6) / (0.1)²
E_net = 54 × 10³ / 10^-2
E_net = 5.4 × 10⁶ N/C towards B.

Q3. Four point charges 2 μC, -5 μC, 2 μC, -5 μC are located at corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the center?

Solution:
By symmetry:
Force due to A (2 μC) cancels Force due to C (2 μC) [Equal and Opposite].
Force due to B (-5 μC) cancels Force due to D (-5 μC).
Net force on center charge = 0 N.

Q4. An electric dipole with dipole moment 4 × 10^-9 C m is aligned at 30° with a uniform electric field of magnitude 5 × 10⁴ N/C. Calculate the torque.

Solution:
τ = pE sinθ
τ = (4 × 10^-9) × (5 × 10⁴) × sin(30°)
τ = 20 × 10^-5 × 0.5
τ = 10 × 10^-5 = 10^-4 N m.

Q5. A point charge of 10 μC is at a distance 5 cm directly above the center of a square of side 10 cm. Find the electric flux magnitude through the square.

Solution:
Imagine a cube of side 10 cm enclosing the charge at its center.
Total flux through cube = q / ε₀.
The square is one face of this cube.
Flux through square = (1/6) × Total Flux
Φ = (1/6) × (10 × 10^-6) / (8.85 × 10^-12)
Φ ≈ 1.88 × 10⁵ N m²/C.

Q6. A pith ball of mass 9 × 10^-5 kg carries a charge of 5 μC. What must be the magnitude of the electric field acting vertically upwards to balance its weight?

Solution:
For equilibrium: Electric Force = Weight
qE = mg
E = mg / q
E = (9 × 10^-5 × 10) / (5 × 10^-6)
E = 90 / 0.5 = 180 N/C.

Q7. Two charges +4e and +e are separated by distance a. Where should a third charge q be placed for it to be in equilibrium?

Solution:
Let distance from +e be x.
Force equality: k(4e)q / (a-x)² = k(e)q / x²
Taking square root: 2 / (a-x) = 1 / x
2x = a - x
3x = a → x = a/3
Position: At distance a/3 from +e charge.

Q8. Calculate the work done in rotating a dipole of dipole moment 3 × 10^-8 C m from stable equilibrium to unstable equilibrium in a field of 10⁴ N/C.

Solution:
Stable (θ₁ = 0°) to Unstable (θ₂ = 180°).
W = pE (cosθ₁ - cosθ₂)
W = pE (cos 0° - cos 180°) = pE (1 - (-1)) = 2pE
W = 2 × 3 × 10^-8 × 10⁴
W = 6 × 10^-4 J.

Q9. An electron falls through a distance of 1.5 cm in a uniform electric field of 2 × 10⁴ N/C. Calculate the time of fall. (Ignore gravity).

Solution:
Acceleration a = eE/m
s = 1/2 at² → t = √(2s/a) = √(2sm/eE)
s = 1.5 × 10^-2 m
t = √( [2 × 1.5×10^-2 × 9.1×10^-31] / [1.6×10^-19 × 2×10⁴] )
t ≈ 2.9 × 10^-9 s.

Q10. A large plane sheet has charge density 8.85 × 10^-8 C/m². Find E-field near it.

Solution:
E = σ / 2ε₀
E = (8.85 × 10^-8) / (2 × 8.85 × 10^-12)
E = 10^-8 / (2 × 10^-12)
E = 10⁴ / 2 = 5000 N/C.

Important Formulae

1. Coulomb's Law

F = k |q₁q₂| / r²

k = 9 × 10⁹ N m² C^-2
In medium: F' = F / K

2. Electric Field Intensity

Point Charge:

E = kQ / r²

Infinite Line Charge:

E = λ / 2πε₀r

Infinite Sheet:

E = σ / 2ε₀

3. Electric Dipole

Dipole Moment:

p = q × 2a

Torque:

τ = pE sinθ

Work Done (Rotation):

W = pE (cosθ₁ - cosθ₂)

4. Gauss's Law

Φ = q_enclosed / ε₀

Φ = Electric Flux (N m² C^-1)
ε₀ = 8.85 × 10^-12 C² N^-1 m^-2

20 NEET Golden Facts

1. Charge on Body: Always an integral multiple of e. Charge less than e (quarks) is not stable in free state.
2. Repulsion is Sure Test: Attraction can happen between charged and neutral body (induction), but repulsion only between like charges.
3. Gold Leaf Electroscope: Used to detect presence of charge and polarity. Divergence is proportional to charge magnitude.
4. Sharp Points: Charge density σ ∝ 1/Radius of curvature. Sharp points leak charge (Corona Discharge).
5. Shell Shielding: Electric field inside a hollow conductor is ALWAYS zero, independent of external fields (Electrostatic Shielding).
6. Dielectric Breakdown: Air becomes conducting if E > 3 × 10⁶ V/m.
7. Null Point: For like charges, null point is between them (closer to smaller charge). For unlike charges, it is outside (closer to smaller charge magnitude).
8. Soap Bubble: When given charge (positive or negative), its radius increases due to outward electrostatic repulsion.
9. Field Lines & Conductor: Field lines always start/end normally (90°) to the surface of a conductor.
10. Stable Equilibrium: Dipole in uniform field is stable when θ = 0° (parallel to E). Unstable when θ = 180°.
11. Flux independent of Shape: Flux through a closed surface depends only on q_in, not on size or shape of surface.
12. Cube Center: Charge at center of cube → Flux = q/ε₀. Flux through ONE face = q/6ε₀.
13. Cube Corner: Charge at corner → Shared by 8 cubes. Flux through cube = q/8ε₀.
14. Force on Dipole: In Uniform E-field, Force=0, Torque≠0. In Non-Uniform E-field, Force≠0, Torque≠0.
15. Electron vs Proton: In same E-field, force is same magnitude (|qE|), but acceleration depends on mass. a_e > a_p.
16. Permittivity: Water has high dielectric constant (K=81), reducing force between ions by 81 times (Reason for solubility).
17. Solid vs Hollow Sphere: For conducting sphere, charge resides on surface. E-field strength vs Distance graph is identical for solid and hollow conducting spheres.
18. Millikan's Drop: Determined quantum of charge e. Condition for suspension: qE = mg.
19. Simple Pendulum: If bob is charged and placed in uniform vertical E-field, effective 'g' changes. g' = g ± qE/m.
20. Gravitational vs Electrostatic: Electrostatic force is much stronger (approx 10³⁶ times) than Gravitational force for electrons.

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