Solution:

We can use Coulomb’s law to solve this problem:

\(F = \frac{1}{4\pi\epsilon_0}\frac{|q_1 \times q_2|}{r^2}\)

Given:

• \(q_1 = 3 \times 10^{-6} C\)
• \(q_2 = -4 \times 10^{-6} C\)
• \(r = 30 \text{ cm} = 0.3 \text{ m}\)

Let’s calculate the magnitude of the force:

\(F = \frac{9 \times 10^9 \times |3 \times 10^{-6} \times (-4 \times 10^{-6})|}{(0.3)^2}\)

\(F = \frac{9 \times 10^9 \times 12 \times 10^{-12}}{0.09}\)

\(F = \frac{108 \times 10^{-3}}{0.09} = 1.2 \text{ N}\)

Since the charges have opposite signs (one positive and one negative), the force will be attractive. The direction of the force on \(q_1\) is toward \(q_2\), and the direction of the force on \(q_2\) is toward \(q_1\).

Therefore, the magnitude of the force is 1.2 N, and it is attractive.

Solution:

The electric field due to a point charge is given by:

\(E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\)

Given:

• \(q = 5 \times 10^{-6} \text{ C}\)
• \(r = 20 \text{ cm} = 0.2 \text{ m}\)

Let’s calculate the electric field:

\(E = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{(0.2)^2}\)

\(E = \frac{45 \times 10^3}{0.04} = 1.125 \times 10^6 \text{ N/C}\)

Since the charge is positive, the electric field points radially outward from the charge.

Therefore, the electric field at the given point is 1.125 × 10^6 N/C directed radially outward from the charge.

Solution:

The capacitance of a parallel plate capacitor is given by:

\(C = \frac{\epsilon_0 \epsilon_r A}{d}\)

Where:

• \(\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}\) (permittivity of free space)
• \(\epsilon_r\) = relative permittivity of the dielectric
• \(A = 400 \text{ cm}^2 = 400 \times 10^{-4} \text{ m}^2 = 4 \times 10^{-2} \text{ m}^2\)
• \(d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}\)

(a) For air (ε_r = 1):

\(C_{air} = \frac{8.85 \times 10^{-12} \times 1 \times 4 \times 10^{-2}}{2 \times 10^{-3}}\)

\(C_{air} = \frac{3.54 \times 10^{-13}}{2 \times 10^{-3}} = 1.77 \times 10^{-10} \text{ F} = 177 \text{ pF}\)

(b) For mica (ε_r = 6):

\(C_{mica} = \frac{8.85 \times 10^{-12} \times 6 \times 4 \times 10^{-2}}{2 \times 10^{-3}}\)

\(C_{mica} = \frac{21.24 \times 10^{-13}}{2 \times 10^{-3}} = 10.62 \times 10^{-10} \text{ F} = 1062 \text{ pF}\)

Therefore, the capacitance is 177 pF with air and 1062 pF with mica as the dielectric material.

Solution:

We can use Gauss’s law to solve this problem. Gauss’s law states:

\(\oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}\)

(a) For r < R (inside the sphere):

Let’s consider a Gaussian sphere of radius r (where r < R) centered at the same point as the charged sphere.

For a uniform volume charge density ρ, the charge enclosed by this Gaussian sphere is:

\(q_{enclosed} = \rho \times \frac{4}{3}\pi r^3\)

Due to symmetry, the electric field must be radially outward and have the same magnitude at all points on the Gaussian sphere. So:

\(E \times 4\pi r^2 = \frac{\rho \times \frac{4}{3}\pi r^3}{\epsilon_0}\)

\(E = \frac{\rho r}{3\epsilon_0}\)

(b) For r > R (outside the sphere):

In this case, the Gaussian sphere of radius r encloses the entire charged sphere. The total charge enclosed is:

\(q_{enclosed} = \rho \times \frac{4}{3}\pi R^3\)

Applying Gauss’s law:

\(E \times 4\pi r^2 = \frac{\rho \times \frac{4}{3}\pi R^3}{\epsilon_0}\)

\(E = \frac{\rho R^3}{3\epsilon_0 r^2}\)

This can be rewritten as:

\(E = \frac{1}{4\pi\epsilon_0} \frac{q_{total}}{r^2}\)

Where \(q_{total} = \frac{4}{3}\pi R^3 \rho\)

Therefore, the electric field is:

(a) For r < R: \(E = \frac{\rho r}{3\epsilon_0}\) (directly proportional to r)

(b) For r > R: \(E = \frac{\rho R^3}{3\epsilon_0 r^2}\) (inversely proportional to r²)

Solution:

Given:

• Dipole moment p = 4 × 10⁻⁹ C·m
• Electric field E = 5 × 10⁵ N/C
• Angle between dipole and field θ = 60°

(a) The torque on an electric dipole in an external electric field is given by:

\(\vec{\tau} = \vec{p} \times \vec{E}\)

The magnitude of the torque is:

\(\tau = p E \sin\theta\)

\(\tau = (4 \times 10^{-9})(5 \times 10^5)\sin 60°\)

\(\tau = (4 \times 10^{-9})(5 \times 10^5)(0.866)\)

\(\tau = 1.732 \times 10^{-3} \text{ N·m}\)

(b) The potential energy of an electric dipole in an external electric field is given by:

\(U = -\vec{p} \cdot \vec{E} = -pE\cos\theta\)

\(U = -(4 \times 10^{-9})(5 \times 10^5)\cos 60°\)

\(U = -(4 \times 10^{-9})(5 \times 10^5)(0.5)\)

\(U = -1 \times 10^{-3} \text{ J}\)

Therefore, the torque on the dipole is 1.732 × 10⁻³ N·m, and the potential energy is -1 × 10⁻³ J.

Solution:

Given that all capacitors have the same value C = 3 μF, let’s analyze this circuit step by step.

First, let’s identify the arrangement:

• C₁ is in series with the rest of the circuit
• C₂ and the combination of C₃ and C₄ are in parallel
• C₃ and C₄ are in series

Step 1: Find the equivalent capacitance of C₃ and C₄ in series.

\(C_{34} = \frac{C_3 \times C_4}{C_3 + C_4} = \frac{3 \times 3}{3 + 3} = \frac{9}{6} = 1.5 \text{ μF}\)

Step 2: Find the equivalent capacitance of C₂ and C₃₄ in parallel.

\(C_{234} = C_2 + C_{34} = 3 + 1.5 = 4.5 \text{ μF}\)

Step 3: Find the equivalent capacitance of C₁ and C₂₃₄ in series.

\(C_{eq} = \frac{C_1 \times C_{234}}{C_1 + C_{234}} = \frac{3 \times 4.5}{3 + 4.5} = \frac{13.5}{7.5} = 1.8 \text{ μF}\)

Therefore, the equivalent capacitance between points A and B is 1.8 μF.

Solution:

This problem requires integrating the contributions from all charge elements along the line.

Consider a small element of charge dq = λ dx at a position x on the x-axis. The distance r from this element to the point P(0, a) is:

\(r = \sqrt{x^2 + a^2}\)

The electric field due to this element at point P has two components – E_x and E_y:

\(dE = \frac{1}{4\pi\epsilon_0}\frac{dq}{r^2}\)

\(dE = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + a^2}\)

From the geometry:

\(dE_x = dE \cdot \frac{x}{r} = \frac{\lambda}{4\pi\epsilon_0}\frac{x dx}{(x^2 + a^2)^{3/2}}\)

\(dE_y = dE \cdot \frac{a}{r} = \frac{\lambda}{4\pi\epsilon_0}\frac{a dx}{(x^2 + a^2)^{3/2}}\)

To find the total electric field, we integrate from x = 0 to x = ∞:

\(E_x = \frac{\lambda}{4\pi\epsilon_0}\int_{0}^{\infty}\frac{x dx}{(x^2 + a^2)^{3/2}}\)

\(E_y = \frac{\lambda}{4\pi\epsilon_0}\int_{0}^{\infty}\frac{a dx}{(x^2 + a^2)^{3/2}}\)

For E_x, we can use the substitution u = x² + a² :

\(E_x = \frac{\lambda}{4\pi\epsilon_0}\int_{a^2}^{\infty}\frac{1}{2u^{3/2}} du = \frac{\lambda}{4\pi\epsilon_0} \left[-\frac{1}{u^{1/2}}\right]_{a^2}^{\infty} = \frac{\lambda}{4\pi\epsilon_0} \cdot \frac{1}{a}\)

For E_y, we can directly integrate:

\(E_y = \frac{\lambda a}{4\pi\epsilon_0}\int_{0}^{\infty}\frac{dx}{(x^2 + a^2)^{3/2}} = \frac{\lambda a}{4\pi\epsilon_0} \left[\frac{x}{a^2\sqrt{x^2 + a^2}}\right]_{0}^{\infty} = \frac{\lambda}{4\pi\epsilon_0} \cdot \frac{1}{a}\)

Therefore, both components are equal:

\(E_x = E_y = \frac{\lambda}{4\pi\epsilon_0 a}\)

The magnitude of the resultant electric field is:

\(E = \sqrt{E_x^2 + E_y^2} = \sqrt{2} \cdot \frac{\lambda}{4\pi\epsilon_0 a}\)

The direction makes an angle of 45° with both the x and y axes (in the first quadrant).

Therefore, the electric field at point P(0, a) has a magnitude of \(\frac{\lambda \sqrt{2}}{4\pi\epsilon_0 a}\) and makes an angle of 45° with the positive x-axis.

Solution:

This problem involves capacitors in series with different dielectrics and the concept of bound charges at dielectric interfaces.

(a) The equivalent capacitance:

We can treat this as three capacitors in series, each with thickness d/3 and filled with different dielectrics:

\(C_1 = \frac{\epsilon_0 \epsilon_1 A}{d/3}\), \(C_2 = \frac{\epsilon_0 \epsilon_2 A}{d/3}\), \(C_3 = \frac{\epsilon_0 \epsilon_3 A}{d/3}\)

For capacitors in series:

\(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}\)

\(\frac{1}{C_{eq}} = \frac{d/3}{\epsilon_0 \epsilon_1 A} + \frac{d/3}{\epsilon_0 \epsilon_2 A} + \frac{d/3}{\epsilon_0 \epsilon_3 A}\)

\(\frac{1}{C_{eq}} = \frac{d}{3\epsilon_0 A} \left(\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} + \frac{1}{\epsilon_3}\right)\)

Therefore:

\(C_{eq} = \frac{3\epsilon_0 A}{d\left(\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} + \frac{1}{\epsilon_3}\right)}\)

(b) Electric field in each dielectric layer:

When capacitors are in series, the charge on each is the same, but the electric fields differ.

The total voltage is:

\(V = E_1 \cdot \frac{d}{3} + E_2 \cdot \frac{d}{3} + E_3 \cdot \frac{d}{3}\)

Using Gauss’s law and the boundary conditions at dielectric interfaces, the displacement field D is constant:

\(D = \epsilon_0 E_1 \epsilon_1 = \epsilon_0 E_2 \epsilon_2 = \epsilon_0 E_3 \epsilon_3\)

Therefore:

\(E_1 = \frac{D}{\epsilon_0 \epsilon_1}\), \(E_2 = \frac{D}{\epsilon_0 \epsilon_2}\), \(E_3 = \frac{D}{\epsilon_0 \epsilon_3}\)

Substituting into the voltage equation:

\(V = \frac{D}{\epsilon_0} \cdot \frac{d}{3} \left(\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} + \frac{1}{\epsilon_3}\right)\)

Solving for D:

\(D = \frac{3\epsilon_0 V}{d\left(\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} + \frac{1}{\epsilon_3}\right)}\)

Therefore, the electric fields are:

\(E_1 = \frac{3V}{d\left(\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} + \frac{1}{\epsilon_3}\right)\epsilon_1}\)

\(E_2 = \frac{3V}{d\left(\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} + \frac{1}{\epsilon_3}\right)\epsilon_2}\)

\(E_3 = \frac{3V}{d\left(\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} + \frac{1}{\epsilon_3}\right)\epsilon_3}\)

(c) Surface charge density at each dielectric interface:

The bound surface charge density at the interface between two dielectrics is given by:

\(\sigma_b = \epsilon_0 (E_2 – E_1)\) for perpendicular fields

At the interface between dielectrics 1 and 2:

\(\sigma_{12} = \epsilon_0 (E_1 – E_2) = \epsilon_0 D \left(\frac{1}{\epsilon_0 \epsilon_1} – \frac{1}{\epsilon_0 \epsilon_2}\right) = D \left(\frac{1}{\epsilon_1} – \frac{1}{\epsilon_2}\right)\)

\(\sigma_{12} = \frac{3\epsilon_0 V}{d\left(\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} + \frac{1}{\epsilon_3}\right)} \left(\frac{1}{\epsilon_1} – \frac{1}{\epsilon_2}\right)\)

At the interface between dielectrics 2 and 3:

\(\sigma_{23} = \epsilon_0 (E_2 – E_3) = D \left(\frac{1}{\epsilon_2} – \frac{1}{\epsilon_3}\right)\)

\(\sigma_{23} = \frac{3\epsilon_0 V}{d\left(\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} + \frac{1}{\epsilon_3}\right)} \left(\frac{1}{\epsilon_2} – \frac{1}{\epsilon_3}\right)\)

Therefore:

(a) The equivalent capacitance is \(\frac{3\epsilon_0 A}{d\left(\frac{1}{\epsilon_1} + \frac{1}{\epsilon_2} + \frac{1}{\epsilon_3}\right)}\)

(b) The electric fields are inversely proportional to the respective dielectric constants

(c) The surface charge densities at the interfaces depend on the differences in reciprocals of the dielectric constants

Solution:

We need to apply the superposition principle to find the net electric field and potential at the center of the square.

Let’s place the square in the xy-plane with its center at the origin and sides parallel to the x and y axes. The four corners have coordinates:

• Corner 1: (a/2, a/2) with charge +q
• Corner 2: (-a/2, a/2) with charge +q
• Corner 3: (-a/2, -a/2) with charge +q
• Corner 4: (a/2, -a/2) with charge -q

Electric Field Calculation:

The distance from each corner to the center is:

\(r = \sqrt{(a/2)^2 + (a/2)^2} = \frac{a}{\sqrt{2}}\)

For each charge, the electric field magnitude at the center is:

\(E_{single} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{q}{a^2/2} = \frac{2q}{4\pi\epsilon_0 a^2}\)

The directions of the electric fields are:

• E₁: from Corner 1 to center (direction: (-1, -1))
• E₂: from Corner 2 to center (direction: (1, -1))
• E₃: from Corner 3 to center (direction: (1, 1))
• E₄: from center to Corner 4 (since negative charge) (direction: (1, -1))

Let’s normalize these direction vectors:

\(\hat{r}_1 = \frac{1}{\sqrt{2}}(-1, -1)\)

\(\hat{r}_2 = \frac{1}{\sqrt{2}}(1, -1)\)

\(\hat{r}_3 = \frac{1}{\sqrt{2}}(1, 1)\)

\(\hat{r}_4 = \frac{1}{\sqrt{2}}(1, -1)\)

The electric field components are:

\(\vec{E}_1 = E_{single} \cdot \hat{r}_1 = \frac{2q}{4\pi\epsilon_0 a^2} \cdot \frac{1}{\sqrt{2}}(-1, -1)\)

\(\vec{E}_2 = E_{single} \cdot \hat{r}_2 = \frac{2q}{4\pi\epsilon_0 a^2} \cdot \frac{1}{\sqrt{2}}(1, -1)\)

\(\vec{E}_3 = E_{single} \cdot \hat{r}_3 = \frac{2q}{4\pi\epsilon_0 a^2} \cdot \frac{1}{\sqrt{2}}(1, 1)\)

\(\vec{E}_4 = -E_{single} \cdot \hat{r}_4 = -\frac{2q}{4\pi\epsilon_0 a^2} \cdot \frac{1}{\sqrt{2}}(1, -1)\)

The net electric field is:

\(\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \vec{E}_4\)

Computing the x-component:

\(E_x = \frac{2q}{4\pi\epsilon_0 a^2} \cdot \frac{1}{\sqrt{2}}(-1 + 1 + 1 – 1) = 0\)

Computing the y-component:

\(E_y = \frac{2q}{4\pi\epsilon_0 a^2} \cdot \frac{1}{\sqrt{2}}(-1 – 1 + 1 + 1) = 0\)

Therefore, the net electric field at the center is zero.

Electric Potential Calculation:

The electric potential due to each charge is:

\(V_{single} = \frac{1}{4\pi\epsilon_0}\frac{q}{r} = \frac{1}{4\pi\epsilon_0}\frac{q}{a/\sqrt{2}} = \frac{\sqrt{2}q}{4\pi\epsilon_0 a}\)

The net potential is the algebraic sum:

\(V_{net} = V_1 + V_2 + V_3 + V_4 = \frac{\sqrt{2}q}{4\pi\epsilon_0 a} + \frac{\sqrt{2}q}{4\pi\epsilon_0 a} + \frac{\sqrt{2}q}{4\pi\epsilon_0 a} – \frac{\sqrt{2}q}{4\pi\epsilon_0 a} = \frac{2\sqrt{2}q}{4\pi\epsilon_0 a}\)

Therefore:

The electric field at the center of the square is zero.

The electric potential at the center is \(\frac{2\sqrt{2}q}{4\pi\epsilon_0 a}\).