A parallel plate capacitor has charge 5×10⁻⁶~C$5\times10^{-6}\mathrm{~C}$. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is 4×10⁻⁶~C$4\times10^{-6}\mathrm{~C}$ then the dielectric constant of the slab is _______. [cite: 183, 184]

Related Formula

The electric field due to a circular arc subtending an angle φ$\phi$ at the center is given by:

Core Logic

Arc AB subtends 90^°$90^\circ$ (one quadrant) at the center. The electric field due to it is given as E$E$. Arc ABC consists of two independent quadrants: arc AB and arc BC. Each quadrant independently creates an electric field of magnitude E$E$ pointing along the bisector of that specific quadrant.

Step 1: Vector Addition

The electric field EAB$\vec{E}_{AB}$ is directed at 45^°$45^\circ$ away from both axes into the third quadrant. The electric field EBC$\vec{E}_{BC}$ is directed at 45^°$45^\circ$ towards the matching opposite quadrant. Since EAB$\vec{E}_{AB}$ and EBC$\vec{E}_{BC}$ are perpendicular to each other, their resultant magnitude is: Eₙₑₜ = √(E² + E²) = √(2)E$$E_{\text{net}} = \sqrt{E^2 + E^2} = \sqrt{2}E$$

Pattern Recognition

Symmetric components of a ring create orthogonal vector fields. Each 90^°$90^\circ$ arc produces a field of magnitude E$E$ directed along its angular bisector. Two adjacent quadrants have bisectors separated by 90^°$90^\circ$, hence use orthogonal vector addition.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Capacitance modification by dielectric: C' = K · C$C' = K \cdot C$ Series combination:

Parallel combination:

Core Logic

Initial capacitance value C = 5$C = 5\ \mu\text{F}$ for all. After dielectric insertion into C₁$C_1$, its value becomes:

The values for the others remain constant:

Step 1: Circuit Topology Analysis

From the network layout, C₁$C_1$ and C₂$C_2$ are configured in a series arm, which is collectively in parallel with C₃$C_3$. Equivalent of the series arm:

Adding the parallel branch C₃$C_3$:

Pattern Recognition

Identify layout components systematically. Series components simplify via product-over-sum, then combine linearly with parallel components.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Principle of superposition for electric fields:

Electric field due to an infinite uniformly charged sheet:

Electric field outside a uniformly charged spherical body:

Core Logic

• Between the sheets (Points A$A$ and B$B$): The two identical infinite parallel sheets produce equal and oppositely directed electric fields that cancel completely (Esheets = 0$\vec{E}_{\text{sheets}} = \vec{0}$). Thus, the net electric field is determined entirely by the charged sphere. Because point A$A$ is located closer to the center of the sphere than point B$B$ (rA < rB$r_A < r_B$):

• Outside the sheets (Points C$C$ and D$D$): The planar electric fields reinforce each other in the same direction (E = (σ)/(ε₀)$E = \frac{\sigma}{\varepsilon_0}$), while the sphere contributes a radial field. Because points C$C$ and D$D$ lie at different radial distances from the center of the sphere (rD < rC$r_D < r_C$), the net field magnitudes differ:

Therefore, the valid relation is EC ≠ ED; EA > EB$\vec{E}_C \neq \vec{E}_D; \vec{E}_A > \vec{E}_B$.

Pattern Recognition

Between two identical positively charged parallel sheets, the sheet fields cancel out completely, making the sphere the sole field source. Outside both sheets, the sheet fields reinforce each other, and asymmetric radial distances to the sphere guarantee unequal net field values.

Evaluation Rubric / Model Answer

Option C: EC ≠ ED; EA > EB$\vec{E}_{C} \neq \vec{E}_{D}; \vec{E}_{A} > \vec{E}_{B}$

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Electric field due to an infinite non-conducting sheet:

Torque on an electric dipole:

where p = qd$p = qd$ is the magnitude of the electric dipole moment.

Core Logic

Given parameters:

• Charge magnitude, q = 2 = 2 × 10⁻⁶ C$q = 2\ \mu\text{C} = 2 \times 10^{-6}\text{ C}$
• Separation length, d = 20 cm = 0.2 m$d = 20\text{ cm} = 0.2\text{ m}$
• Surface charge density, σ = 100 /m² = 100 × 10⁻⁶ C/m²$\sigma = 100\ \mu\text{C/m}^2 = 100 \times 10^{-6}\text{ C/m}^2$
• Orientation angle with field, θ = 30°$\theta = 30^{\circ}$
• Permittivity of free space, ε₀ = 8.85 × 10⁻¹² C²/(N ²)$\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2/(\text{N}\cdot\text{m}^2)$

Step 1: Compute Electric Field and Net Torque

The electric field generated by the infinite non-conducting sheet is uniform:

The dipole moment is:

Substitute p$p$, E$E$, and θ$\theta$ into the torque formula:

Pattern Recognition

Equal and opposite charges on a rigid rod constitute an electric dipole. In a uniform field, the net translational force vanishes (Fₙₑₜ = 0$\vec{F}_{\text{net}} = \vec{0}$), leaving only a pure restoring torque τ = pE θ$\tau = pE\sin\theta$.

Evaluation Rubric / Model Answer

Option B: 1.12 Nm

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Equivalent capacitance for capacitors in parallel:

Core Logic

By tracing electric potential along the ideal connecting wires, label the two plates of each capacitor with their connected terminal (A$A$ or B$B$).

Redrawing the circuit reveals that all 4 capacitors are connected directly in parallel across terminals A$A$ and B$B$.

Step 1: Calculate Equivalent Value

Since all four identical capacitors are connected in parallel:

Given each capacitor has C = 16$C = 16\,\mu\text{F}$:

Pattern Recognition

Bridging wires across alternate terminals typically simplifies an apparent ladder into a pure parallel arrangement. Always apply the node-potential labeling method first to identify terminals sharing identical potentials.

Evaluation Rubric / Model Answer

64

Chapter Mix

Class 12 Physics: Electrostatics