Problem: Two point charges of +3 μC and -5 μC are placed 15 cm apart in air. Calculate the magnitude and direction of the electrostatic force on each charge. What happens to the force if the charges are immersed in water (dielectric constant κ = 80)?

Solution 2 Marks

Given:
$q_1 = +3 \times 10^{-6}\, \text{C},\; q_2 = -5 \times 10^{-6}\, \text{C}$
$r = 15\, \text{cm} = 0.15\, \text{m}$
$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\, \text{Nm}^2\text{C}^{-2}$

Step 1: Force in air (vacuum)
Using Coulomb’s law:
$F = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2} = 9 \times 10^9 \times \frac{|3 \times 10^{-6} \times (-5) \times 10^{-6}|}{(0.15)^2}$
$F = 9 \times 10^9 \times \frac{15 \times 10^{-12}}{0.0225} = 9 \times 10^9 \times 6.67 \times 10^{-10} = 6.0\, \text{N}$

Step 2: Direction and medium effect
Since charges are opposite, force is attractive.
Magnitude of force on each charge = 6.0 N (Newton’s third law)
In water: $F_{\text{water}} = \frac{F_{\text{air}}}{\kappa} = \frac{6.0}{80} = 0.075\, \text{N}$

$\boxed{F_{\text{air}} = 6.0\,\text{N (attractive)}, \quad F_{\text{water}} = 0.075\,\text{N}}$

2. Net Force on Charge Due to Multiple Charges

Problem: Three charges +2 μC, -3 μC, and +4 μC are placed at the vertices of an equilateral triangle of side 10 cm. Calculate the magnitude and direction of the net force on the +2 μC charge.

Three charges at vertices of equilateral triangle showing force vectors — Forces on +2 μC charge due to other two charges

Solution 3 Marks

Given:
$q_A = +2\,\mu\text{C},\; q_B = -3\,\mu\text{C},\; q_C = +4\,\mu\text{C}$
Side length $a = 0.1\, \text{m}$

Step 1: Force due to -3 μC charge
$F_{AB} = 9 \times 10^9 \times \frac{|2 \times 10^{-6} \times (-3) \times 10^{-6}|}{(0.1)^2} = 5.4\, \text{N}$
Direction: Attractive → towards B

Step 2: Force due to +4 μC charge
$F_{AC} = 9 \times 10^9 \times \frac{|2 \times 10^{-6} \times 4 \times 10^{-6}|}{(0.1)^2} = 7.2\, \text{N}$
Direction: Repulsive → away from C

Step 3: Vector addition
Angle between forces = 60° (equilateral triangle)
$F_{\text{net}} = \sqrt{F_{AB}^2 + F_{AC}^2 + 2F_{AB}F_{AC}\cos 60^\circ}$
$F_{\text{net}} = \sqrt{5.4^2 + 7.2^2 + 2(5.4)(7.2)(0.5)} = \sqrt{29.16 + 51.84 + 38.88} = \sqrt{119.88} = 10.95\, \text{N}$

$\boxed{10.95\,\text{N at an angle of } 33.7^\circ \text{ from } F_{AC}}$

3. Electric Field Due to Point Charges

Problem: Two point charges +10 nC and -10 nC are placed 8 cm apart. Calculate the electric field at a point 5 cm from each charge (forming an equilateral triangle).

Solution 2 Marks

Given:
$q_1 = +10^{-8}\, \text{C},\; q_2 = -10^{-8}\, \text{C},\; r = 0.05\, \text{m}$

Step 1: Field magnitudes
$E_1 = \frac{1}{4\pi\epsilon_0} \frac{q_1}{r^2} = 9 \times 10^9 \times \frac{10^{-8}}{(0.05)^2} = 3.6 \times 10^4\, \text{N/C}$ (away from +q)
$E_2 = 9 \times 10^9 \times \frac{10^{-8}}{(0.05)^2} = 3.6 \times 10^4\, \text{N/C}$ (towards -q)

Step 2: Vector addition
Angle between fields = 60°
Vertical components cancel (equal magnitude, opposite direction)
Horizontal components add: $E_{\text{net}} = 2E_1 \cos 30^\circ = 2 \times 3.6 \times 10^4 \times \frac{\sqrt{3}}{2} = 6.24 \times 10^4\, \text{N/C}$
Direction: Horizontal, from +q towards -q

$\boxed{6.24 \times 10^4\,\text{N/C horizontally from +q to -q}}$

4. Electric Field on Dipole Axis

Problem: An electric dipole with charges ±5 nC separated by 2 mm is placed in vacuum. Calculate the electric field at a point 15 cm from the dipole center (a) on the axial line, and (b) on the equatorial line.

Solution 3 Marks

Given:
$q = 5 \times 10^{-9}\, \text{C},\; 2a = 2 \times 10^{-3}\, \text{m} \Rightarrow a = 10^{-3}\, \text{m}$
$r = 0.15\, \text{m},\; \frac{1}{4\pi\epsilon_0} = 9 \times 10^9$

Step 1: Dipole moment
$p = q \times 2a = 5 \times 10^{-9} \times 2 \times 10^{-3} = 10^{-11}\, \text{C·m}$

Step 2: Axial field (r >> a)
$E_{\text{axial}} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} = 9 \times 10^9 \times \frac{2 \times 10^{-11}}{(0.15)^3} = 9 \times 10^9 \times \frac{2 \times 10^{-11}}{3.375 \times 10^{-3}} = 53.33\, \text{N/C}$
Direction: Same as dipole moment (–q to +q)

Step 3: Equatorial field
$E_{\text{eq}} = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} = \frac{53.33}{2} = 26.67\, \text{N/C}$
Direction: Opposite to dipole moment

$\boxed{\text{(a) } 53.33\,\text{N/C} \quad \text{(b) } 26.67\,\text{N/C}}$

5. Torque on Dipole in Uniform Field

Problem: An electric dipole of dipole moment 4 × 10⁻⁹ C·m is placed at 30° with a uniform electric field of magnitude 5 × 10⁴ N/C. Calculate (a) the torque acting on the dipole, and (b) the potential energy of the dipole.

Solution 2 Marks

Given:
$p = 4 \times 10^{-9}\, \text{C·m},\; E = 5 \times 10^4\, \text{N/C},\; \theta = 30^\circ$

Part (a): Torque
$\tau = pE \sin\theta = (4 \times 10^{-9}) \times (5 \times 10^4) \times \sin 30^\circ$
$\tau = 20 \times 10^{-5} \times 0.5 = 10^{-4}\, \text{N·m}$
Direction: Perpendicular to plane containing p and E (right-hand rule)

Part (b): Potential energy
$U = -pE \cos\theta = -(4 \times 10^{-9}) \times (5 \times 10^4) \times \cos 30^\circ$
$U = -20 \times 10^{-5} \times \frac{\sqrt{3}}{2} = -1.732 \times 10^{-4}\, \text{J}$

$\boxed{\text{(a) } 10^{-4}\,\text{N·m} \quad \text{(b) } -1.732 \times 10^{-4}\,\text{J}}$

6. Electric Flux Through a Surface

Problem: A uniform electric field of magnitude 3 × 10³ N/C makes an angle of 60° with the normal to a square surface of side 10 cm. Calculate the electric flux through the surface.

Solution 1 Mark

Given:
$E = 3 \times 10^3\, \text{N/C},\; a = 0.1\, \text{m},\; \theta = 60^\circ$

Step 1: Area calculation
$A = a^2 = (0.1)^2 = 0.01\, \text{m}^2$

Step 2: Flux calculation
$\phi = \vec{E} \cdot \vec{A} = EA \cos\theta = (3 \times 10^3) \times 0.01 \times \cos 60^\circ$
$\phi = 30 \times 0.5 = 15\, \text{N·m}^2\text{/C}$

$\boxed{15\,\text{N·m}^2\text{/C}}$

7. Gauss’s Law: Field Due to Infinite Line Charge

Problem: An infinitely long straight wire has a uniform linear charge density of 2 μC/m. Calculate the electric field at a perpendicular distance of 5 cm from the wire.

Gaussian cylindrical surface around infinite line charge — Gaussian cylindrical surface for infinite line charge

Solution 2 Marks

Given:
$\lambda = 2 \times 10^{-6}\, \text{C/m},\; r = 0.05\, \text{m},\; \epsilon_0 = 8.854 \times 10^{-12}\, \text{C}^2\text{N}^{-1}\text{m}^{-2}$

Step 1: Apply Gauss’s law
For cylindrical Gaussian surface of radius r and length l:
Flux through curved surface: $\phi = E \times 2\pi r l$
Charge enclosed: $q = \lambda l$

Step 2: Solve for E
By Gauss’s law: $E \times 2\pi r l = \frac{\lambda l}{\epsilon_0}$
$E = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{2 \times 10^{-6}}{2\pi \times 8.854 \times 10^{-12} \times 0.05}$
$E = \frac{2 \times 10^{-6}}{2.78 \times 10^{-12}} = 7.2 \times 10^5\, \text{N/C}$
Direction: Radially outward (since λ > 0)

$\boxed{7.2 \times 10^5\,\text{N/C radially outward}}$

8. Gauss’s Law: Charged Spherical Shell

Problem: A thin spherical shell of radius 10 cm has a uniform surface charge density of 80 μC/m². Calculate the electric field at distances (a) 5 cm, (b) 10 cm, and (c) 20 cm from the center of the shell.

Solution 3 Marks

Given:
$R = 0.1\, \text{m},\; \sigma = 80 \times 10^{-6}\, \text{C/m}^2$
Total charge: $q = \sigma \times 4\pi R^2 = 80 \times 10^{-6} \times 4\pi \times (0.1)^2 = 1.005 \times 10^{-5}\, \text{C}$

Part (a): r = 5 cm (inside)
For points inside a charged shell: $E = 0$

Part (b): r = 10 cm (on surface)
$E = \frac{1}{4\pi\epsilon_0} \frac{q}{R^2} = 9 \times 10^9 \times \frac{1.005 \times 10^{-5}}{(0.1)^2} = 9.045 \times 10^6\, \text{N/C}$

Part (c): r = 20 cm (outside)
$E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} = 9 \times 10^9 \times \frac{1.005 \times 10^{-5}}{(0.2)^2} = 2.26 \times 10^6\, \text{N/C}$

$\boxed{\text{(a) } 0 \quad \text{(b) } 9.045 \times 10^6\,\text{N/C} \quad \text{(c) } 2.26 \times 10^6\,\text{N/C}}$

9. Field Due to Infinite Plane Sheet

Problem: Two large parallel metal plates carry surface charge densities of +17.7 × 10⁻²² C/m² and -17.7 × 10⁻²² C/m² on their inner faces. Calculate the electric field (a) between the plates, and (b) outside the plates.

Solution 2 Marks

Given:
$\sigma = 17.7 \times 10^{-22}\, \text{C/m}^2,\; \epsilon_0 = 8.854 \times 10^{-12}\, \text{C}^2\text{N}^{-1}\text{m}^{-2}$

Step 1: Field due to single sheet
For an infinite sheet: $E_{\text{sheet}} = \frac{\sigma}{2\epsilon_0}$
$E_{\text{sheet}} = \frac{17.7 \times 10^{-22}}{2 \times 8.854 \times 10^{-12}} = 10^{-10}\, \text{N/C}$

Step 2: Superposition
(a) Between plates: Fields add up
$E_{\text{between}} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0} = 2 \times 10^{-10}\, \text{N/C}$
Direction: From positive to negative plate
(b) Outside plates: Fields oppose and cancel
$E_{\text{outside}} = \frac{\sigma}{2\epsilon_0} - \frac{\sigma}{2\epsilon_0} = 0$

$\boxed{\text{(a) } 2 \times 10^{-10}\,\text{N/C} \quad \text{(b) } 0}$

10. Charge Quantization Problem

Problem: A polythene piece rubbed with wool acquires a charge of -3 × 10⁻⁷ C. (a) Estimate the number of electrons transferred. (b) Is there any transfer of mass from wool to polythene? If yes, calculate the mass transferred. (Mass of electron = 9.1 × 10⁻³¹ kg)

Solution 2 Marks

Given:
$q = -3 \times 10^{-7}\, \text{C},\; e = -1.6 \times 10^{-19}\, \text{C},\; m_e = 9.1 \times 10^{-31}\, \text{kg}$

Part (a): Number of electrons
Since charge is negative, electrons are transferred FROM wool TO polythene.
$n = \frac{|q|}{|e|} = \frac{3 \times 10^{-7}}{1.6 \times 10^{-19}} = 1.875 \times 10^{12}$ electrons

Part (b): Mass transfer
Yes, mass is transferred with electrons.
Mass transferred: $\Delta m = n \times m_e = (1.875 \times 10^{12}) \times (9.1 \times 10^{-31}) = 1.706 \times 10^{-18}\, \text{kg}$

$\boxed{\text{(a) } 1.875 \times 10^{12} \text{ electrons (wool → polythene)} \quad \text{(b) } 1.706 \times 10^{-18}\,\text{kg}}$

Exam Tip: For numerical problems on Electric Charges and Fields:

• Always convert units to SI before calculations (cm → m, μC → C)
• For vector quantities (force, field), determine both magnitude AND direction
• Apply superposition principle carefully for multiple charges
• For Gauss’s law problems, choose Gaussian surface matching symmetry (sphere, cylinder, pillbox)
• Remember: Field inside conductor = 0 in electrostatic conditions; field inside charged shell = 0

← Back to Theory Next: Electrostatic Potential →

10 Solved Numericals on Electric Charges and Fields

10 Solved Numericals on Electric Charges and Fields

1. Force Between Two Point Charges

2. Net Force on Charge Due to Multiple Charges

3. Electric Field Due to Point Charges

4. Electric Field on Dipole Axis

5. Torque on Dipole in Uniform Field

6. Electric Flux Through a Surface

7. Gauss’s Law: Field Due to Infinite Line Charge

8. Gauss’s Law: Charged Spherical Shell

9. Field Due to Infinite Plane Sheet

10. Charge Quantization Problem

Editorial Team

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1. Force Between Two Point Charges

2. Net Force on Charge Due to Multiple Charges

3. Electric Field Due to Point Charges

4. Electric Field on Dipole Axis

5. Torque on Dipole in Uniform Field

6. Electric Flux Through a Surface

7. Gauss’s Law: Field Due to Infinite Line Charge

8. Gauss’s Law: Charged Spherical Shell

9. Field Due to Infinite Plane Sheet

10. Charge Quantization Problem

Editorial Team

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