CUET UG Physics Electrostatics Electric Field and Field Lines Dipole Continuous Charge Distribution

Must-Know (15–20 detailed bullets)

• Electric field due to a point charge ( q ) at distance ( r ): ( E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} ), direction radially outward for positive ( q ). Example: ( q = 1.6 \times 10^{-19} \, \text{C} ), ( r = 5.3 \times 10^{-11} \, \text{m} ) (Bohr radius), ( E \approx 5.14 \times 10^{11} \, \text{N/C} ).
• Electric field lines start on positive charges and end on negative charges; never form closed loops.
• Number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.
• Electric field inside a conductor is zero in electrostatic conditions; excess charge resides on the surface.
• Electric dipole moment ( \vec{p} = q \times 2\vec{a} ), where ( 2a ) is the separation between ( +q ) and ( -q ); SI unit: C·m. Example: ( q = 10^{-6} \, \text{C}, 2a = 2 \, \text{cm} \Rightarrow p = 2 \times 10^{-8} \, \text{C·m} ).
• Electric field on the axial line of a dipole at distance ( r ) from center: ( E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3} ), parallel to ( \vec{p} ).
• Electric field on the equatorial line of a dipole at distance ( r ) from center: ( E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} ), opposite to ( \vec{p} ).
• Torque on a dipole in uniform electric field: ( \vec{\tau} = \vec{p} \times \vec{E} ), magnitude ( \tau = pE \sin\theta ); zero when ( \vec{p} \parallel \vec{E} ).
• Potential energy of dipole in electric field: ( U = -\vec{p} \cdot \vec{E} = -pE \cos\theta ); minimum at ( \theta = 0^\circ ), maximum at ( \theta = 180^\circ ).
• Electric field due to a uniformly charged infinite straight wire: ( E = \frac{1}{4\pi\varepsilon_0} \frac{2\lambda}{r} ), where ( \lambda ) is linear charge density, ( r ) is perpendicular distance.
• Electric field due to uniformly charged infinite plane sheet: ( E = \frac{\sigma}{2\varepsilon_0} ), independent of distance; ( \sigma ) = surface charge density.
• Electric field due to uniformly charged thin spherical shell:
• Outside (( r > R )): ( E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} )
• Inside (( r < R )): ( E = 0 )
• On surface: ( E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2} )
• Electric field due to uniformly charged solid sphere (volume charge density ( \rho )):
• Outside (( r > R )): ( E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} )
• Inside (( r < R )): ( E = \frac{1}{4\pi\varepsilon_0} \frac{Qr}{R^3} )
• Electric field is discontinuous across a charged surface; discontinuity ( = \frac{\sigma}{\varepsilon_0} ).
• For a dipole, electric field at general point: ( E = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} \sqrt{3\cos^2\theta + 1} ), where ( \theta ) is angle from dipole axis.
• Electric field due to a ring of charge ( Q ), radius ( R ), at axial distance ( x ): ( E = \frac{1}{4\pi\varepsilon_0} \frac{Qx}{(x^2 + R^2)^{3/2}} ); maximum at ( x = R/\sqrt{2} ).
• Electric field due to a uniformly charged disc at axial point: ( E = \frac{\sigma}{2\varepsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) ); as ( R \to \infty ), becomes field of infinite plane.
• Electric field is a vector quantity; superposition principle applies: ( \vec{E}_{\text{net}} = \sum \vec{E}_i ) or ( \int d\vec{E} ) for continuous distribution.
• Linear charge density ( \lambda = \frac{dq}{dl} ), surface charge density ( \sigma = \frac{dq}{dA} ), volume charge density ( \rho = \frac{dq}{dV} ).
• Electric field lines never intersect; closer lines indicate stronger field.

Difficulty Level

Intermediate — requires understanding of vector fields, symmetry, and integration for continuous distributions, but core formulas are derivable from NCERT.

Common CUET Traps

• Trap: Assuming electric field inside a charged shell is ( \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} ) like a point charge.
  Avoid: Remember ( E = 0 ) inside a uniformly charged thin spherical shell (verify from NCERT).

• Trap: Using ( E = \frac{\sigma}{\varepsilon_0} ) for a single infinite sheet.
  Avoid: Correct formula is ( E = \frac{\sigma}{2\varepsilon_0} ); ( \frac{\sigma}{\varepsilon_0} ) is for field between plates of a parallel plate capacitor.

• Trap: Thinking torque on dipole is maximum when aligned with field.
  Avoid: Torque ( \tau = pE \sin\theta ), so maximum at ( \theta = 90^\circ ), zero at ( \theta = 0^\circ ).

Practice MCQs

Q1. The electric field at a distance ( r ) (where ( r \gg 2a )) on the equatorial line of a dipole of moment ( p ) is:
A. ( \frac{1}{4\pi\varepsilon_0} \frac{p}{r^2} )
B. ( \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^2} )
C. ( \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} )
D. ( \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3} )
Answer: C
Explanation: Electric field on equatorial line is ( E = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} ).
Why others fail: Option D is the axial field, often confused with equatorial.

Q2. A uniformly charged thin spherical shell has total charge ( Q ) and radius ( R ). The electric field at a point ( r = R/2 ) from center is:
A. ( \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2} )
B. ( \frac{1}{4\pi\varepsilon_0} \frac{4Q}{R^2} )
C. 0
D. ( \frac{1}{4\pi\varepsilon_0} \frac{Q}{4R^2} )
Answer: C
Explanation: Electric field inside a uniformly charged thin spherical shell is zero.
Why others fail: Option A is field on surface, mistakenly used for inside.

Q3. The torque on an electric dipole of moment ( \vec{p} ) placed in a uniform electric field ( \vec{E} ) is maximum when the angle between ( \vec{p} ) and ( \vec{E} ) is:
A. 0°
B. 90°
C. 180°
D. 45°
Answer: B
Explanation: Torque ( \tau = pE \sin\theta ), maximum at ( \theta = 90^\circ ).
Why others fail: Students confuse with potential energy, which is maximum at 180°.

Q4. A long straight wire carries a linear charge density ( \lambda = 2 \times 10^{-6} \, \text{C/m} ). The electric field at a perpendicular distance of 0.5 m is (use ( \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 )):
A. ( 7.2 \times 10^4 \, \text{N/C} )
B. ( 3.6 \times 10^4 \, \text{N/C} )
C. ( 1.8 \times 10^4 \, \text{N/C} )
D. ( 9 \times 10^3 \, \text{N/C} )
Answer: A
Explanation: ( E = \frac{1}{4\pi\varepsilon_0} \frac{2\lambda}{r} = 9 \times 10^9 \times \frac{2 \times 2 \times 10^{-6}}{0.5} = 7.2 \times 10^4 \, \text{N/C} ).
Why others fail: Option B omits the factor 2, common mistake in formula recall.

Q5. Two infinite parallel sheets have uniform surface charge densities ( +\sigma ) and ( -\sigma ). The electric field in the region between them is:
A. ( \frac{\sigma}{2\varepsilon_0} )
B. ( \frac{\sigma}{\varepsilon_0} )
C. 0
D. ( \frac{2\sigma}{\varepsilon_0} )
Answer: B
Explanation: Fields add constructively: ( \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} ).
Why others fail: Option A is field due to single sheet, not considering superposition.

Last‑Minute Revision

• ⚠️ Dipole moment ( \vec{p} = q \times 2\vec{a} ), direction –ve to +ve charge.
• ⚠️ ( E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3} ), ( E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} ).
• ⚠️ Torque ( \tau = pE \sin\theta ), not ( pE \cos\theta ).
• ⚠️ Potential energy ( U = -pE \cos\theta ), minimum at ( \theta = 0^\circ ).
• ⚠️ Field due to infinite wire: ( E = \frac{2k\lambda}{r} ).
• ⚠️ Field due to infinite sheet: ( E = \frac{\sigma}{2\varepsilon_0} ).
• ⚠️ Field between capacitor plates: ( E = \frac{\sigma}{\varepsilon_0} ).
• ⚠️ Inside charged shell: ( E = 0 ).
• ⚠️ Outside solid sphere: ( E \propto \frac{1}{r^2} ).
• ⚠️ Inside solid sphere: ( E \propto r ).
• ⚠️ Field lines never intersect — always.
• ⚠️ Field lines start on +q, end on –q.
• ⚠️ Number of field lines ∝ charge magnitude.
• ⚠️ ( \lambda = dq/dl ), ( \sigma = dq/dA ), ( \rho = dq/dV ).
• ⚠️ Superposition applies to electric field (vector sum).
• ⚠️ At ( x = R/\sqrt{2} ), field on axis of ring is maximum.
• ⚠️ For dipole, ( E \propto 1/r^3 ) at large distances.
• ⚠️ Field discontinuity across surface: ( \Delta E = \sigma / \varepsilon_0 ).