Electric Charges and Fields Chapter-Wise Test 3

Let \( x \) be distance from \( +10 \, \mu\text{C} \), then \( 0.5 - x \) from \( -5 \, \mu\text{C} \).

\( \frac{10 \times 10^{-6}}{x^2} = \frac{5 \times 10^{-6}}{(0.5 - x)^2} \), \( 10 (0.5 - x)^2 = 5 x^2 \).

\( 2 (0.25 - x + x^2) = x^2 \), \( 0.5 - 2 x + 2 x^2 = x^2 \).

\( x^2 - 2 x + 0.5 = 0 \), \( x = \frac{2 \pm \sqrt{4 - 2}}{2} = \frac{2 \pm \sqrt{2}}{2} \).

\( x = 1 - \frac{\sqrt{2}}{2} \approx 0.293 \, \text{m} \) (between charges).

For a dipole, the net charge is zero, and the field results from the combined effect of two equal and opposite charges. At large distances, the fields from each charge partially cancel, leading to a faster fall-off (\( 1/r^3 \)) compared to the \( 1/r^2 \) dependence of a single charge.

\( \phi = \frac{q}{\varepsilon_0} \).

\( q = \phi \varepsilon_0 = 5.65 \times 10^5 \times 8.854 \times 10^{-12} = 5 \times 10^{-6} \, \text{C} = 5 \, \mu\text{C} \).

Let \( x \) be distance from \( +8 \, \mu\text{C} \), then \( 0.2 - x \) from \( -2 \, \mu\text{C} \).

\( \frac{8 \times 10^{-6}}{x^2} = \frac{2 \times 10^{-6}}{(0.2 - x)^2} \), \( 8 (0.2 - x)^2 = 2 x^2 \).

\( 4 (0.04 - 0.4 x + x^2) = x^2 \), \( 0.16 - 1.6 x + 4 x^2 = x^2 \).

\( 3 x^2 - 1.6 x + 0.16 = 0 \), \( x = \frac{1.6 \pm \sqrt{1.6^2 - 4 \times 3 \times 0.16}}{6} = \frac{1.6 \pm 1.2}{6} \).

\( x = 0.133 \, \text{m} \) (between charges).

The field depends only on the surface charge density (\( \sigma/2\varepsilon_0 \)), as derived from Gauss’s law for an infinite plane. The material affects charge distribution, but for a given \( \sigma \), the field is universal, assuming infinite extent and uniformity.

High charge density produces a stronger electric field. The convention of field lines dictates that their closeness (density) reflects field strength, so they cluster near regions where the charge per unit area or volume is greater.

Dipole moment: \( p = q \times 2a = 7 \times 10^{-6} \times 3 \times 10^{-3} = 2.1 \times 10^{-8} \, \text{C m} \).

Torque: \( \tau = p E \sin \theta = 2.1 \times 10^{-8} \times 4 \times 10^4 \times \sin 60^\circ = 8.4 \times 10^{-4} \times \frac{\sqrt{3}}{2} = 7.28 \times 10^{-4} \, \text{N m} \).

In a uniform field, the forces on the positive and negative charges of the dipole are equal in magnitude and opposite in direction, resulting in no net force. However, these forces act at different points, creating a torque that tends to align the dipole with the field.

\( E = \frac{k q}{r^2} \).

\( 5 \times 10^3 = 9 \times 10^9 \times \frac{q}{(0.4)^2} \).

\( q = \frac{5 \times 10^3 \times 0.16}{9 \times 10^9} = 8.89 \times 10^{-8} \, \text{C} \).

Midpoint distance = 35 cm = 0.35 m.

\( E_1 = 9 \times 10^9 \times \frac{9 \times 10^{-6}}{(0.35)^2} = 6.61 \times 10^5 \, \text{N/C} \) (towards \( -3 \, \mu\text{C} \)).

\( E_2 = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{(0.35)^2} = 2.2 \times 10^5 \, \text{N/C} \) (towards \( -3 \, \mu\text{C} \)).

Net \( E = 6.61 \times 10^5 + 2.2 \times 10^5 = 8.81 \times 10^5 \, \text{N/C} \).

Surface area: \( A = 4 \pi r^2 = 4 \pi (0.23)^2 = 0.2116 \pi \, \text{m}^2 \).

Charge: \( q = \sigma A = 35 \times 10^{-6} \times 0.2116 \pi = 2.33 \times 10^{-5} \, \text{C} \).

\( E = \frac{k |q|}{r^2} \).

\( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), \( q = 3 \times 10^{-6} \, \text{C} \), \( r = 3 \, \text{m} \).

\( E = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{(3)^2} = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{9} = 3 \times 10^3 \, \text{N/C} \).

Total flux: \( \phi = \frac{q}{\varepsilon_0} = \frac{3 \times 10^{-6}}{8.854 \times 10^{-12}} = 3.39 \times 10^5 \, \text{Nm}^2/\text{C} \).

Flux per face (6 faces): \( \phi_{\text{face}} = \frac{3.39 \times 10^5}{6} = 5.65 \times 10^4 \, \text{Nm}^2/\text{C} \).

Total charge is the algebraic sum (additivity of charges).

\( q_{\text{total}} = (+3 \times 10^{-6}) + (-5 \times 10^{-6}) + (+2 \times 10^{-6}) = 3 - 5 + 2 = 0 \, \mu\text{C} \).

\( E_1 = \frac{k |q_1|}{r^2} = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{(0.2)^2} = 9 \times 10^5 \, \text{N/C} \) (away).

\( E_2 = 9 \times 10^9 \times \frac{8 \times 10^{-6}}{(0.2)^2} = 1.8 \times 10^6 \, \text{N/C} \) (towards).

Angle between \( E_1 \) and \( E_2 \) is 60°. Net \( E = \sqrt{E_1^2 + E_2^2 + 2 E_1 E_2 \cos 60^\circ} \).

\( E = \sqrt{(9 \times 10^5)^2 + (1.8 \times 10^6)^2 + 2 \times 9 \times 10^5 \times 1.8 \times 10^6 \times 0.5} = 2.29 \times 10^6 \, \text{N/C} \).