NEET Physics: Electric Charges and Fields
Electrostatics forms the bedrock of Classical Electromagnetism and constitutes 8% to 10% of the NEET Physics paper (typically 2–3 questions annually). Every concept below is distilled with exact mathematical rigor, high-yield traps, and memory shortcuts.
1. Electric Charge: Fundamental Properties & Charging Methods
Electric charge is an intrinsic scalar property of elementary particles that gives rise to electrostatic interactions. The SI unit of charge is the Coulomb (C), and its dimensional formula is [A T].
Three Fundamental Properties of Electric Charge
- Additivity of Charge: Total electric charge of an isolated system is the algebraic sum of all individual charges: Qnet = Σ qi = q1 + q2 + ... + qn. Signs (+ or −) must be included algebraically.
- Conservation of Charge: The total charge of an isolated system remains strictly constant over time. Charge cannot be created or destroyed, only transferred from one body to another (e.g., in pair production γ → e− + e+, net charge remains 0).
- Quantization of Charge: Any physically observable charge exists only in integral multiples of the elementary charge unit e = 1.6 × 10−19 C:q = ± n e (where n = 1, 2, 3, ...)
NEET TRAP 1 — Mass Change During Charging:
When a neutral body acquires a negative charge (−q), it gains electrons, so its mass slightly increases by Δm = n me = (qe) me. Conversely, when it is positively charged, its mass decreases. Here me = 9.1 × 10−31 kg.
Methods of Charging
- Charging by Friction: Rubbing two insulators transfers electrons from the body with lower work function to the one with higher work function. Both bodies acquire equal and opposite charges.
- Charging by Conduction: Direct physical contact between a charged body and an uncharged conductor. When two identical conducting spheres of charges q1 and q2 are brought in contact and separated, the final charge on each is:q' = q1 + q22
- Charging by Induction: Charging without direct physical contact. A charged rod brought near a grounded conductor induces an opposite charge on the conductor:q' = − q [1 − 1K]For a metal (conductor), K = ∞ &implies; q' = −q (100% induced charge).
Golden Rule: Repulsion is the SURE Test of Electrification!
An uncharged body can be attracted by a charged body due to electrostatic induction. Therefore, attraction alone does NOT prove that a body is charged. Only repulsion between two bodies confirms that both carry like charges.
2. Coulomb's Law & Principle of Superposition
Coulomb's Law quantifies the electrostatic force between two stationary point charges in a vacuum or medium.
Scalar and Vector Form
The magnitude of electrostatic force between two point charges q1 and q2 separated by distance r in vacuum is:
F0 = 14πε0 |q1 q2|r2 = k |q1 q2|r2
where k = 14πε0 ≈ 8.9875 × 109 N·m2/C2 ≈ 9 × 109 N·m2/C2, and the permittivity of free space is ε0 = 8.854 × 10−12 C2/(N·m2) = F/m.
In vector form, the force on charge 2 exerted by charge 1 is:
F21 = 14πε0 q1 q2|r2 − r1|3 (r2 − r1) = − F12
Coulomb forces obey Newton's Third Law and are central, conservative forces.
Figure 1: Coulomb's Law in vector notation showing action-reaction pair F₁₂ = -F₂₁, position vectors, and dielectric reduction.
Effect of Dielectric Medium (Relative Permittivity K)
When the space between charges is filled with a dielectric medium of relative permittivity εr (also known as dielectric constant K), the permittivity becomes ε = K ε0:
Fmed = 14πε |q1 q2|r2 = F0K
NEET TRAP 2 — Effective Air Distance for Partially Filled Medium:
If a dielectric slab of thickness t and dielectric constant K is placed between two charges separated by total distance r, the effective distance in air is:
reff = r − t + t √K
and the force becomes:
F = 14πε0 q1 q2(r − t + t √K)2
3. Electric Field & Electric Field Lines
The electric field E at a point in space is defined as the electrostatic force experienced by a vanishingly small positive test charge q0 placed at that point:
E = Fq0 [Unit: N/C or V/m; Dimensions: M L T−3 A−1]
Electric Field due to a Point Charge
E = 14πε0 |q|r2 = k qr2
Direction: Radially outward for a positive source charge (+q); radially inward for a negative source charge (−q).
Properties of Electric Field Lines & NEET Traps
- Field lines originate from positive charges and terminate on negative charges (or extend to infinity).
- The tangent to a field line at any point gives the direction of the net electric field vector E at that point.
- Two field lines can NEVER intersect! If they intersected, there would be two tangents (two directions of E) at one point, which is physically impossible.
- Field lines do not form closed loops because the electrostatic field is conservative (∮ E · dl = 0). (Unlike magnetic field lines which form closed loops).
- Field lines are always perpendicular to the surface of a conductor in electrostatic equilibrium.
- There are no electric field lines inside the material of a conductor in electrostatic equilibrium (Ein = 0).
- The relative density of field lines in a region is directly proportional to the magnitude of the electric field (crowded lines &implies; stronger field).
4. Electric Dipole: Field, Torque & Potential Energy
An electric dipole consists of two equal and opposite point charges (+q and −q) separated by a small distance 2a.
The Electric Dipole Moment vector p is directed strictly from negative charge to positive charge:
p = q (2a) [Unit: C·m; Dimensions: [L T A]]
Figure 2: Electric dipole field lines, axial vs equatorial field calculations, and torque in a uniform external field.
Electric Field of a Short Dipole (r ≫ a)
| Position | Vector Formula | Magnitude Formula | Direction relative to p |
|---|---|---|---|
| Axial Line (End-on) | Eax = 2 k pr3 | Eax = 2 k pr3 | Parallel to p (θ = 0°) |
| Equatorial Line (Broadside-on) | Eeq = − k pr3 | Eeq = k pr3 | Antiparallel to p (θ = 180°) |
| General Point (r, θ) | — | E = k pr3 √3 cos2θ + 1 | tan α = 12 tan θ |
Golden Ratio for NEET:
For equal distances r from the center of a short dipole:
Eaxial = 2 × Eequatorial
Note that for a dipole, E ∝ 1r3, whereas for a single point charge, E ∝ 1r2.Dipole in an External Electric Field
- In Uniform Electric Field:
- Net force: Fnet = (+qE) + (−qE) = 0. (No translational motion).
- Net torque: τ = p × E &implies; τ = p E sin θ. Maximum torque occurs at θ = 90° (τmax = p E).
- Electrostatic Potential Energy: U = − p · E = − p E cos θ.
- Work done in rotating dipole from angle θ1 to θ2:W = ΔU = p E (cos θ1 − cos θ2)
- Stable Equilibrium: θ = 0° (p parallel to E). Here τ = 0 and Umin = −p E.
- Unstable Equilibrium: θ = 180° (p antiparallel to E). Here τ = 0 and Umax = +p E.
- In Non-Uniform Electric Field: Both net force and net torque are non-zero: Fnet ≠ 0 and τ ≠ 0. The dipole experiences both translation and rotation!
5. Electric Flux & Gauss's Law
Electric flux ΦE represents the total number of electric field lines passing normally through a given surface area A:
ΦE = ∫ E · dA = ∫ E dA cos θ [Unit: N·m2/C or V·m; Dimensions: M L3 T−3 A−1]
Here θ is the angle between the electric field E and the outward normal area vector dA.
Gauss's Theorem
The total electric flux through any closed Gaussian surface in vacuum is equal to 1ε0 times the net charge enclosed inside that surface:
Φclosed = ∮ E · dA = qenclosedε0
Figure 3: Gauss's Law concept, enclosed vs external charge, and the classic NEET Cube and curved surface traps.
MASTER TABLE: NEET CUBE SYMMETRY TRAPS
Consider a charge q placed at different positions relative to a cube of side length a:
- Charge at Center of Cube:
Total flux through the cube: Φtotal = qε0
Flux through each of the 6 identical faces: Φface = q6ε0 - Charge at Center of One Face:
Symmetry requires 2 identical cubes to completely enclose the charge.
Total flux through the given cube: Φcube = q2ε0
Flux through the contact face = 0 (field lines are parallel to this face).
Flux through each of the remaining 5 faces: Φeach = q10ε0 - Charge at One Corner (Vertex) of the Cube:
Symmetry requires 8 identical cubes sharing that corner to completely enclose the charge.
Total flux through the given cube: Φcube = q8ε0
Flux through the 3 adjacent faces sharing that corner = 0 (field lines graze these surfaces).
Flux through each of the 3 opposite faces: Φopp = q24ε0 - Charge at Center of One Edge:
Symmetry requires 4 identical cubes meeting at that edge.
Total flux through the given cube: Φcube = q4ε0
6. Applications of Gauss's Law (Standard Geometries)
Figure 4: Gaussian surfaces for infinitely long wire, non-conducting infinite sheet, conducting sheet, spherical shell, and E vs r graphs.
1. Field due to Infinitely Long Straight Uniformly Charged Wire
For a wire with uniform linear charge density λ = qL, using a co-axial cylindrical Gaussian surface of radius r and length L:
E = λ2πε0r = 2 k λr &implies; E ∝ 1r
2. Field due to Uniformly Charged Infinite Non-Conducting Plane Sheet
For a thin sheet with surface charge density σ = qA, using a pillbox cylinder of cross-sectional area A penetrating both sides:
2 E A = σ Aε0 &implies; E = σ2ε0
NEET TRAP 3 — Sheet vs Conducting Plate:
- Thin non-conducting sheet: E = σ2ε0 (charge exists on one layer).
- Conducting plate / sheet: Charges reside on both surfaces with density σ, or field inside conductor is zero, so:Econducting = σε0 = 2 × Enon-conducting
- Both fields are independent of the distance r from the sheet!
3. Field due to Uniformly Charged Thin Spherical Shell (Radius R, Total Charge Q)
- Inside the Shell (r < R):
Since the Gaussian surface inside encloses zero charge (qenc = 0):
Ein = 0 (Electrostatic Shielding) - At the Surface (r = R):Esurface = 14πε0 QR2 = σε0 (Maximum Field)
- Outside the Shell (r > R):Eout = 14πε0 Qr2 ∝ 1r2(The entire charge appears to be concentrated at the center).
7. Motion of a Charged Particle in Uniform Electric Field
Consider a particle of mass m and charge q projected horizontally with velocity v0 into a region between two parallel plates of length L having a uniform vertical electric field E.
- Horizontal motion: No horizontal force &implies; ax = 0 &implies; x = v0 t &implies; t = xv0.
- Vertical motion: Constant electric force Fy = q E &implies; ay = q Em.
- Trajectory equation:y = 12 ay t2 = 12 (q Em) (xv0)2 = q E2 m v02 x2 (Parabolic Path)
- Vertical Deflection at exit (x = L):y = q E L22 m v02 ∝ qm
NEET TRAP 4 — Specific Charge Deflection:
When different charged particles (electron, proton, alpha particle) enter the field with the same kinetic energy K = 12 m v02:
y = q E L24 K ∝ q
The deflection depends purely on the charge q, NOT the mass! But if they enter with the same velocity, deflection is proportional to specific charge qm. Read question wording with extreme care!Master Formulas & 20 Golden NEET Facts
Rapid-revision summary covering every formula, constant, unit, and high-probability trap tested in NEET (Physics: Electric Charges and Fields).
Part 1: The Master Formula Matrix
| Concept / Entity | Formula | SI Unit / Dimension | Key Conditions & Notes |
|---|---|---|---|
| Quantization of Charge | q = ± n e | Coulomb (C) • [A T] | e = 1.6 × 10−19 C; n ∈ ℕ. Valid microscopically. |
| Coulomb's Law (Vacuum) | F0 = 14πε0 |q1 q2|r2 | Newton (N) • [M L T−2] | 14πε0 = 9 × 109 N·m2/C2. Point charges at rest. |
| Coulomb's Law (Medium) | Fmed = F0K = F0εr | Newton (N) | K ≥ 1 (Kair ≈ 1, Kwater = 80, Kmetal = ∞). |
| Partially Filled Medium | F = 14πε0 |q1 q2|(r − t + t√K)2 | Newton (N) | Slab of thickness t and constant K between charges. |
| Electric Field of Point Charge | E = k qr2 | N/C or V/m • [M L T−3 A−1] | Radially outward for +q, inward for −q. |
| Electric Dipole Moment | p = q (2a) | C·m • [L T A] | Vector directed strictly from −q to +q. |
| Dipole Axial Field | Eax = 2 k pr3 | N/C | For short dipole (r ≫ a). In same direction as p. |
| Dipole Equatorial Field | Eeq = k pr3 | N/C | Antiparallel to p. Ratio: Eax = 2 Eeq. |
| Dipole General Point Field | E = k pr3 √3 cos2θ + 1 | N/C | Direction: tan α = 12 tan θ. |
| Torque on Dipole | τ = p × E &implies; τ = p E sin θ | N·m • [M L2 T−2] | Maximum at θ = 90°; zero at θ = 0°, 180°. |
| Dipole Potential Energy | U = − p · E = − p E cos θ | Joule (J) • [M L2 T−2] | Stable: θ = 0° (U = −p E). Unstable: θ = 180° (U = +p E). |
| Work to Rotate Dipole | W = p E (cos θ1 − cos θ2) | Joule (J) | From 0° → 90°: W = p E. From 0° → 180°: W = 2 p E. |
| Electric Flux | ΦE = ∫ E · dA = E A cos θ | N·m2/C or V·m • [M L3 T−3 A−1] | Scalar quantity. θ is angle with surface normal. |
| Gauss's Law | Φ = ∮ E · dA = qencε0 | N·m2/C | Depends ONLY on net enclosed charge. |
| Infinite Line Charge | E = λ2πε0r = 2 k λr | N/C | E ∝ 1r; λ = qL (Linear charge density). |
| Infinite Non-Conducting Sheet | E = σ2ε0 | N/C | Independent of distance r; σ = qA. |
| Infinite Conducting Plate | E = σε0 | N/C | Double that of non-conducting sheet. Independent of r. |
| Charged Spherical Shell | r < R: E = 0 r = R: E = k QR2 = σε0 r > R: E = k Qr2 | N/C | Zero field inside (electrostatic shielding). Maximum at surface. |
| Particle Deflection in Uniform E | y = q E L22 m v02 = q E L24 K | Meter (m) | Deflection y ∝ qm (same v0); y ∝ q (same K). |
Part 2: 20 Golden NEET Must-Know Facts & Traps
FACT 01
Charge is Invariant Under Relativistic Speeds
Unlike mass (m = m0√1 − v2/c2), the electric charge of an electron or proton is strictly independent of speed and reference frame. Specific charge (q/m), however, decreases as speed approaches c because mass increases.
TRAP 02
Repulsion vs Attraction Diagnostic
A charged body can attract a neutral body through electrostatic induction. Therefore, attraction is an ambiguous indicator. Repulsion is the ONLY foolproof test of electrification between two charged bodies.
FACT 03
Superposition Principle: Pairwise Immunity
The presence of a third or fourth charge does NOT change the electrostatic force between two existing charges. It only alters the net resultant vector force on any one charge. Always calculate pairwise forces independently!
TRAP 04
Dielectric Constant of Conductors
For vacuum, K = 1; for air, K ≈ 1.0006; for water, K = 80; but for any perfect conductor (metal), K = ∞. Hence, the electrostatic force between two charges embedded inside a metal conductor drops to strictly zero!
FACT 05
Collinear 3-Charge System Equilibrium
If two identical charges +Q are placed at distance 2d apart, a central charge q placed at the midpoint is in equilibrium for any value. However, for the ENTIRE SYSTEM to be in equilibrium, the central charge must be:
q = − Q4
Moreover, this entire-system equilibrium is unstable.
TRAP 06
Earnshaw's Theorem & Equilibrium Traps
It is mathematically impossible to achieve stable electrostatic equilibrium for a system of stationary charges using inverse-square electrostatic forces alone. If equilibrium is stable for longitudinal displacement, it is always unstable for transverse displacement.
FACT 07
Neutral Point Formula (Shortcut for NEET)
The distance x of the zero-electric-field point from the smaller charge q1 (where charges are separated by distance d):
- For like charges (+q1 and +q2): x = d√q2 / q1 + 1 (located between the charges).
- For unlike charges (+q1 and −q2 with q1 < q2): x = d√q2 / q1 − 1 (outside, on the side of the smaller charge).
FACT 08
Electric Field Lines Never Form Closed Loops
Electrostatic fields are conservative, meaning ∮ E · dl = 0. Therefore, lines of electrostatic force cannot form closed loops (unlike induced electric fields or magnetic field lines).
TRAP 09
Field Lines Normal to Conducting Surfaces
In electrostatic equilibrium, electric field lines must intersect a conducting surface at exactly 90°. If there were a tangential component Et, surface electrons would experience force q Et and flow, violating static equilibrium.
FACT 10
Electric Dipole Convention Difference
In Physics, the dipole moment p points strictly from negative charge (−q) to positive charge (+q). In Chemistry, it is drawn from positive to negative. On the NEET Physics exam, always use −q → +q!
FACT 11
Dipole Golden Ratio: Axial vs Equatorial
For any short dipole at the same distance r:
EaxialEequatorial = 2 & E ∝ 1r3
Note: Point charge falls off as 1/r2; Dipole falls off as 1/r3; Quadrupole falls off as 1/r4.
TRAP 12
Dipole in Uniform vs Non-Uniform Field
- In a uniform electric field: Fnet = 0, but τ may be non-zero &implies; Pure rotation, NO linear acceleration.
- In a non-uniform electric field: Both Fnet ≠ 0 and τ ≠ 0 &implies; Translates AND rotates simultaneously!
FACT 13
Dipole Work Done in Rotation
To rotate an electric dipole in an external field E from stable equilibrium (θ = 0°):
- Rotation by 90°: W = p E (1 − cos 90°) = p E.
- Rotation by 180° (to unstable equilibrium): W = p E (1 − cos 180°) = 2 p E.
TRAP 14
Gauss's Law Surface Field vs Enclosed Charge
In the formula ∮ E · dA = qencε0:
- The electric field E at any point on the Gaussian surface is the resultant field created by ALL charges (both inside and outside).
- The total flux Φ, however, depends ONLY on charges enclosed inside.
FACT 15
Cube Corner Flux Trap
When a point charge q is placed at one vertex (corner) of a cube:
- Total flux through the cube = q8ε0.
- Flux through the 3 faces meeting at that corner = 0 (field lines lie in these planes, E ⊥ dA).
- Flux through each of the 3 opposite faces = 13 (q8ε0) = q24ε0.
FACT 16
Cube Face-Center Flux Trap
When a point charge q is at the center of one face of a cube:
- Total flux through the cube = q2ε0 (needs 2 cubes).
- Flux through the shared face = 0.
- Flux through each of the other 5 faces = q10ε0.
TRAP 17
Non-Conducting vs Conducting Sheet Field
- Thin non-conducting plane sheet: E = σ2ε0.
- Conducting plate / sheet: Charges reside on two opposing faces, so E = σε0.
- Notice that both are completely independent of distance r!
FACT 18
Charged Spherical Shell Internal Field is Zero
Inside a uniformly charged spherical shell or hollow conductor, Ein = 0 everywhere. A sensitive instrument placed inside a metal cavity is fully shielded from external fields — this is Electrostatic Shielding (why staying inside a car during lightning is safe).
TRAP 19
Deflection in Transverse E Field: Velocity vs Kinetic Energy
When particles enter a uniform transverse electric field:
- If projected with the same speed (v0): y ∝ qm (electron deflects far more than proton).
- If projected with the same kinetic energy (K): y ∝ q (electron and proton suffer identical deflection magnitude!).
FACT 20
Concentric Spherical Shells Charge Induction
When an inner conducting sphere has charge +q, it induces −q on the inner surface of an outer concentric shell, and +q on its outer surface. If the outer shell is grounded, the outer surface charge flows to ground, leaving only the inner induced −q!
NEET Physics: 8 Solved Numericals (3-Box Pedagogy)
Each problem below represents a standard, highly-repeated NEET question type solved via our proven 3-Box Pedagogy: (1) Given & Concept → (2) Step-by-Step Mathematical Calculation → (3) NEET Speed Shortcut.
Problem 1: Charge Sharing & Coulomb's Force
NEET Core
Two identical small conducting spheres A and B carry charges +q and −3q and are separated by a distance r, experiencing an attractive force F. A third identical neutral sphere C is first touched to sphere A, then touched to sphere B, and finally removed. What is the new electrostatic force between A and B in terms of F?
Box 1: Given & Core Concept
- Initial charges: qA = +q, qB = −3q, separation = r.
- Initial force magnitude: F = k |q × (−3q)|r2 = 3 k q2r2.
- Conduction rule: When two identical spheres of charges q1 and q2 touch, each ends with q1 + q22.
Box 2: Step-by-Step Mathematical Calculation
1. Sphere C (neutral, charge = 0) touches sphere A (+q):
q'A = q'C = q + 02 = q2
2. Sphere C (now with charge q/2) touches sphere B (−3q):
q'B = q2 + (−3q)2 = −5q22 = −5q4
3. Calculate new force F' between A and B at distance r:
F' = k |q'A × q'B|r2 = k (q2) × (5q4)r2 = 58 k q2r2
4. Express F' in terms of F:
Since k q2r2 = F3 &implies; F' = 58 (F3) = 5F24 (Repulsive force, as both became negative/positive respectively: A is positive, B is negative &implies; wait: q'A = +q/2 and q'B = −5q/4, so opposite signs &implies; Attractive).
Final Answer: F' = 524 F (Attractive).
Final Answer: F' = 524 F (Attractive).
Box 3: NEET Speed Shortcut / Pro Tip
Remember the general pattern when identical C touches A (q1) then B (q2):
q'A = q12 and q'B = q1 + 2q24.
Direct substitution: q'B = q + 2(−3q)4 = −5q4. Takes under 15 seconds!
q'A = q12 and q'B = q1 + 2q24.
Direct substitution: q'B = q + 2(−3q)4 = −5q4. Takes under 15 seconds!
Problem 2: Vector Superposition on Equilateral Triangle
Vector Mechanics
Three equal positive point charges +Q are placed at the three vertices of an equilateral triangle of side a. What is the magnitude of the net electrostatic force on any one of the charges?
Box 1: Given & Core Concept
- Side length = a; vertex charges = +Q, +Q, +Q.
- Each charge experiences two repulsive forces of equal magnitude F0 = k Q2a2 from the other two charges.
- Angle between the two force vectors = 60° (vertically opposite angle of the 60° vertex).
- Vector addition: Fnet = √F12 + F22 + 2 F1 F2 cos θ.
Box 2: Step-by-Step Mathematical Calculation
1. Compute magnitude of individual pairwise forces:
F1 = F2 = F0 = 14πε0 Q2a2
2. Apply vector resultant formula with θ = 60° (where cos 60° = 12):
Fnet = √F02 + F02 + 2 F02 (12) = √3 F02 = √3 F0
3. Substitute F0:
Fnet = √3 k Q2a2 = √3 Q24πε0 a2
Direction: Along the angle bisector pointing away from the triangle's centroid.
Direction: Along the angle bisector pointing away from the triangle's centroid.
Box 3: NEET Speed Shortcut / Pro Tip
Memorize the standard resultant of two equal vectors F0:
• θ = 60° &implies; Fnet = √3 F0
• θ = 90° &implies; Fnet = √2 F0 (Square vertex)
• θ = 120° &implies; Fnet = F0
• θ = 60° &implies; Fnet = √3 F0
• θ = 90° &implies; Fnet = √2 F0 (Square vertex)
• θ = 120° &implies; Fnet = F0
Problem 3: Zero Electric Field (Neutral Point)
Electric Field
Two point charges +q and +9q are separated by a distance of 40 cm in air. At what distance from the smaller charge +q is the electric field zero?
Box 1: Given & Core Concept
- q1 = +q (smaller charge), q2 = +9q. Total separation d = 40 cm.
- Since both charges are like charges (both positive), the neutral point lies between them on the line joining them.
- Condition for zero field: E1 = E2.
Box 2: Step-by-Step Mathematical Calculation
1. Let the neutral point be at distance x from +q. Then its distance from +9q is (d − x):
k qx2 = k (9q)(d − x)2
2. Cancel k q from both sides and take square root:
1x = 3d − x
3. Cross-multiply and solve for x:
d − x = 3x &implies; 4x = d &implies; x = d4
Substituting d = 40 cm:
x = 404 = 10 cm from the charge +q (and 30 cm from +9q).
Substituting d = 40 cm:
x = 404 = 10 cm from the charge +q (and 30 cm from +9q).
Box 3: NEET Speed Shortcut / Pro Tip
Use the direct Golden Formula:
x = d√q2q1 + 1 = 40√9 + 1 = 403 + 1 = 10 cm.
Solves in 5 seconds without quadratic equations!
Problem 4: Electric Dipole Torque & Rotation Work
Dipole Dynamics
An electric dipole consists of two opposite charges of magnitude 2 μC separated by 2 cm. It is placed in a uniform electric field of 105 N/C at an angle of 30° with the field. Calculate: (a) The torque experienced by the dipole, and (b) The work required to rotate the dipole from stable equilibrium (θ = 0°) to unstable equilibrium (θ = 180°).
Box 1: Given & Core Concept
- q = 2 × 10−6 C, separation 2a = 2 × 10−2 m.
- E = 105 N/C; θ = 30°.
- Dipole moment: p = q(2a) = (2 × 10−6) × (2 × 10−2) = 4 × 10−8 C·m.
- Torque formula: τ = p E sin θ. Work formula: W = p E (cos θ1 − cos θ2).
Box 2: Step-by-Step Mathematical Calculation
1. Torque calculation at θ = 30° (where sin 30° = 0.5):
τ = p E sin 30° = (4 × 10−8) × 105 × 0.5 = 2 × 10−3 N·m
2. Work done to rotate from θ1 = 0° to θ2 = 180°:
W = p E [cos 0° − cos 180°] = p E [1 − (−1)] = 2 p E
W = 2 × (4 × 10−8) × 105 = 8 × 10−3 J = 8 mJ.
W = 2 × (4 × 10−8) × 105 = 8 × 10−3 J = 8 mJ.
Box 3: NEET Speed Shortcut / Pro Tip
Direct Work Shortcuts from stable state (θ = 0°):
• Rotate by 90° &implies; W = pE
• Rotate by 180° (inversion) &implies; W = 2pE
Here pE = 4 × 10−3 J &implies; W = 2 × 4 × 10−3 = 8 mJ. Done instantly!
• Rotate by 90° &implies; W = pE
• Rotate by 180° (inversion) &implies; W = 2pE
Here pE = 4 × 10−3 J &implies; W = 2 × 4 × 10−3 = 8 mJ. Done instantly!
Problem 5: Gauss's Law Flux through Cube Faces
Electric Flux
A point charge q = 8.85 μC is placed at one corner (vertex) of a cube of edge length 10 cm. Calculate: (a) The total electric flux passing through the cube, and (b) The flux through each of the three opposite faces not meeting at this corner. (Take ε0 = 8.85 × 10−12 C2/(N·m2)).
Box 1: Given & Core Concept
- q = 8.85 × 10−6 C; ε0 = 8.85 × 10−12 F/m.
- Symmetry principle: 8 identical cubes sharing that vertex are needed to enclose charge q symmetrically.
- Total flux through one cube: Φcube = q8ε0.
- The 3 faces meeting at that corner have Φ = 0 because E is parallel to those faces (E ⊥ dA).
- The remaining flux is divided equally among the 3 opposite faces: Φopp = 13 Φcube = q24ε0.
Box 2: Step-by-Step Mathematical Calculation
1. Total flux through the given cube:
Φcube = q8ε0 = 8.85 × 10−68 × (8.85 × 10−12) = 1068 = 1.25 × 105 N·m2/C
2. Flux through each of the 3 opposite faces:
Φopp = Φcube3 = q24ε0 = 10624 ≈ 4.17 × 104 N·m2/C
Box 3: NEET Speed Shortcut / Pro Tip
Notice the edge length (10 cm) is a distractor! Gauss flux is strictly independent of surface dimensions.
When q = 8.85 μC, the factor qε0 simplifies directly to 106!
• Φcube = 106 / 8 = 1.25 × 105.
• Φopp = 106 / 24 = 4.17 × 104.
When q = 8.85 μC, the factor qε0 simplifies directly to 106!
• Φcube = 106 / 8 = 1.25 × 105.
• Φopp = 106 / 24 = 4.17 × 104.
Problem 6: Line Charge & Orbiting Electron
Gauss Law Mechanics
An infinite line charge produces an electric field of magnitude 9 × 104 N/C at a perpendicular distance of 2 cm. (a) Find its linear charge density λ. (b) If an electron (m = 9.1 × 10−31 kg, e = 1.6 × 10−19 C) revolves in a circular path of radius 2 cm around this line, find its orbital velocity.
Box 1: Given & Core Concept
- E = 9 × 104 N/C; r = 2 cm = 0.02 m.
- Electric field of line charge: E = 2 k λr &implies; λ = E r2 k.
- Orbital motion: Electrostatic attractive force provides necessary centripetal force: m v2r = e E &implies; v = √e E rm.
Box 2: Step-by-Step Mathematical Calculation
1. Linear charge density λ:
λ = (9 × 104) × 0.022 × (9 × 109) = 180018 × 109 = 10−7 C/m = 0.1 μC/m
2. Orbital velocity of revolving electron:
v2 = e E rm = (1.6 × 10−19) × (9 × 104) × 0.029.1 × 10−31 = 2.88 × 10−169.1 × 10−31 ≈ 3.16 × 1014
v = √3.16 × 1014 ≈ 1.78 × 107 m/s.
v = √3.16 × 1014 ≈ 1.78 × 107 m/s.
Box 3: NEET Speed Shortcut / Pro Tip
Notice that since E ∝ 1/r, the product E × r = 2kλ is independent of radius!
Thus, the orbital speed v = √2 k e λm is constant at ALL radii! (NEET conceptual question favorite: orbital speed does NOT depend on r).
Thus, the orbital speed v = √2 k e λm is constant at ALL radii! (NEET conceptual question favorite: orbital speed does NOT depend on r).
Problem 7: Specific Charge Deflection in Electric Field
Particle Dynamics
An electron enters horizontally with speed v0 = 2 × 107 m/s into a uniform vertical electric field E = 1820 V/m between two parallel plates of length L = 10 cm. Find the vertical deflection of the electron at the exit. (e = 1.6 × 10−19 C, me = 9.1 × 10−31 kg).
Box 1: Given & Core Concept
- v0 = 2 × 107 m/s; L = 0.1 m; E = 1820 N/C.
- Time inside plates: t = Lv0.
- Vertical acceleration: ay = e Eme.
- Vertical deflection: y = 12 ay t2 = e E L22 me v02.
Box 2: Step-by-Step Mathematical Calculation
1. Time of flight inside the plates:
t = 0.12 × 107 = 5 × 10−9 s = 5 ns
2. Vertical acceleration of electron:
ay = (1.6 × 10−19) × 18209.1 × 10−31 = 2.912 × 10−169.1 × 10−31 = 3.2 × 1014 m/s2
3. Vertical deflection y:
y = 12 × (3.2 × 1014) × (5 × 10−9)2 = 1.6 × 1014 × 25 × 10−18 = 40 × 10−4 m = 4.0 mm.
Box 3: NEET Speed Shortcut / Pro Tip
Notice the specific charge factor: eme ≈ 1.76 × 1011 C/kg.
Direct formula: y = 12 (em) E (Lv)2 = 0.5 × (1.76 × 1011) × 1820 × (25 × 10−18) = 4.0 mm.
Direct formula: y = 12 (em) E (Lv)2 = 0.5 × (1.76 × 1011) × 1820 × (25 × 10−18) = 4.0 mm.
Problem 8: Concentric Spherical Shells
Electrostatic Shielding
Two thin concentric conducting spherical shells of radii R and 2R carry charges +Q and −3Q respectively. Determine the electric field magnitude at distances: (a) r = 0.5R, (b) r = 1.5R, and (c) r = 3R from the common center.
Box 1: Given & Core Concept
- Inner shell: radius R, charge +Q.
- Outer shell: radius 2R, charge −3Q.
- Gauss's law for concentric spheres: E(r) = 14πε0 qenc(r)r2.
Box 2: Step-by-Step Mathematical Calculation
1. Region (a) r = 0.5R (inside both shells):
Gaussian sphere of radius 0.5R encloses zero charge: qenc = 0.
Ea = 0 (Electrostatic Shielding).
Ea = 0 (Electrostatic Shielding).
2. Region (b) r = 1.5R (between inner and outer shells):
Gaussian sphere encloses only the inner shell charge (+Q):
Eb = k Q(1.5R)2 = k Q2.25 R2 = 4 k Q9 R2 (Radially outward).
Eb = k Q(1.5R)2 = k Q2.25 R2 = 4 k Q9 R2 (Radially outward).
3. Region (c) r = 3R (outside both shells):
Gaussian sphere encloses both charges: qenc = (+Q) + (−3Q) = −2Q.
|Ec| = k |−2Q|(3R)2 = 2 k Q9 R2 (Radially inward towards center).
|Ec| = k |−2Q|(3R)2 = 2 k Q9 R2 (Radially inward towards center).
Box 3: NEET Speed Shortcut / Pro Tip
For any concentric spherical system:
• Inside innermost shell &implies; ALWAYS E = 0.
• Between shells &implies; outer shell contributes NOTHING! E is due solely to inner shell.
• Outside &implies; Add ALL charges algebraically as a point charge at origin!
• Inside innermost shell &implies; ALWAYS E = 0.
• Between shells &implies; outer shell contributes NOTHING! E is due solely to inner shell.
• Outside &implies; Add ALL charges algebraically as a point charge at origin!
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