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

where, V$V$ = electrostatic potential of a conducting sphere σ$\sigma$ = surface charge density r$r$ = radius of the sphere

Core Logic

For any conducting sphere, the potential on its surface is related to its surface charge density by:

When the two spheres of radii r₁ = R$r_1 = R$ and r₂ = 3R$r_2 = 3R$ are brought into contact, charge flows between them until they reach an identical electric potential:

V₁ = V₂$V_1 = V_2$

Step 1: Ratio Calculation

Equate the potentials of the two spheres after separation:

Pattern Recognition

Sees: "Conducting spheres brought in contact" → Electric potentials become equal: V₁ = V₂$V_1 = V_2$. Shortcut: Since V ∝ σ r$V \propto \sigma r$, equal potential directly implies σ₁ r₁ = σ₂ r₂$\sigma_1 r_1 = \sigma_2 r_2$. Thus, the ratio of final densities is simply the inverse ratio of their radii: (σ₁)/(σ₂) = (r₂)/(r₁) = (3)/(1) = 3$\frac{\sigma_1}{\sigma_2} = \frac{r_2}{r_1} = \frac{3}{1} = 3$. This bypasses computing the individual final charges entirely! ✓

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Core Logic

The dielectrics k₁ = 5$k_1 = 5$ and k₂ = 3$k_2 = 3$ split the capacitor separation horizontally into two layers, each of thickness (d)/(2)$\frac{d}{2}$. Thus, they act as two capacitors connected in series:

• Capacitor 1: C₁ = k₁ (ε₀ A)/(d/2) = 5 × (2 ε₀ A)/(d) = 10 (ε₀ A)/(d)$C_1 = k_1 \frac{\varepsilon_0 A}{d/2} = 5 \times \frac{2 \varepsilon_0 A}{d} = 10 \frac{\varepsilon_0 A}{d}$
• Capacitor 2: C₂ = k₂ (ε₀ A)/(d/2) = 3 × (2 ε₀ A)/(d) = 6 (ε₀ A)/(d)$C_2 = k_2 \frac{\varepsilon_0 A}{d/2} = 3 \times \frac{2 \varepsilon_0 A}{d} = 6 \frac{\varepsilon_0 A}{d}$

Step 1: Compute Base Capacitance Factor

First, find the term (ε₀ A)/(d)$\frac{\varepsilon_0 A}{d}$ in SI units:

• A = 4~cm² = 4 × 10⁻⁴~m²$A = 4\mathrm{~cm}^2 = 4 \times 10^{-4}\mathrm{~m}^2$
• d = 1.77~mm = 1.77 × 10⁻³~m$d = 1.77\mathrm{~mm} = 1.77 \times 10^{-3}\mathrm{~m}$
• ε₀ = 8.85 × 10⁻¹²~F/m$\varepsilon_0 = 8.85 \times 10^{-12}\mathrm{~F/m}$

Now find C₁$C_1$ and C₂$C_2$:

• C₁ = 10 × 2~pF = 20~pF$C_1 = 10 \times 2\mathrm{~pF} = 20\mathrm{~pF}$
• C₂ = 6 × 2~pF = 12~pF$C_2 = 6 \times 2\mathrm{~pF} = 12\mathrm{~pF}$

Calculate their series equivalent Ceq$C_{\text{eq}}$:

Step 2: Add Parallel Capacitor

Another capacitor of Cₚ = 7.5~pF$C_p = 7.5\mathrm{~pF}$ is connected in parallel with the combination:

Pattern Recognition

Sees: Dielectric boundary parallel to the plates → Series capacitors. Trap: Don't treat horizontally split layers as parallel; split in distance d$d$ means series, while split in area A$A$ means parallel. Shortcut: Notice (35.4)/(1.77)$\frac{35.4}{1.77}$ is exactly 20$20$. This makes the numerical calculations incredibly clean! ✓

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Core Logic

An infinite plane sheet produces a uniform electric field E$\vec{E}$ directed normally away from the sheet.

As shown in the image layout, the dipole moment vector p$\vec{p}$ (pointing from -q$-q$ to +q$+q$) is oriented parallel to the electric field vectors E$\vec{E}$:

• Torque evaluation:

• Potential Energy evaluation:

• Net Force evaluation:

Since the electric field is uniform, the force on +q$+q$ balances the force on -q$-q$, meaning Fₙₑₜ = 0$F_{\text{net}} = 0$.

Pattern Recognition

When a dipole aligns perfectly with a uniform electric field (p ∥ E$\vec{p} \parallel \vec{E}$), it reaches stable equilibrium. Stable equilibrium fundamentally means minimum potential energy (U = -pE$U = -pE$) and zero torque.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

where, φ$\phi$ = net electric flux through the closed surface qenclosed$q_{\text{enclosed}}$ = net charge enclosed by the surface ε₀$\epsilon_0$ = permittivity of free space

Core Logic

According to Gauss's Law, the total electric flux through a closed surface depends only on the charge enclosed inside it, completely independent of the radius of the surface.

Rearranging the formula to solve for q$q$:

Substitute the given values:

Pattern Recognition

Distractor alert: The radius (8.0~cm$8.0\mathrm{~cm}$) is extra data meant to mislead. Gauss's flux depends entirely on the magnitude of the enclosed charge, not the physical surface area configuration.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

q = C · Vc$q = C \cdot V_c$

Core Logic

• Initial Charge Calculation:

Before closing the switch, capacitor C₁$C_1$ accumulates total charge:

Uncharged capacitor C₂$C_2$ holds q₂' = 0$q_2' = 0$.

• Redistribution at Equilibrium:

Closing the switch causes parallel redistribution until they reach a common potential Vc$V_c$:

• Final Charges:

Hence, the charges are 10$10\mu\mathrm{C}$ and 20$20\mu\mathrm{C}$ respectively.

Pattern Recognition

For parallel combination setups, the final charge splits in the exact direct ratio of their capacitances: q₁ : q₂ = C₁ : C₂ = 6 : 12 = 1 : 2$q_1 : q_2 = C_1 : C_2 = 6 : 12 = 1 : 2$. Out of 30$30\mu\mathrm{C}$, the breakdown must cleanly be 10$10\mu\mathrm{C}$ and 20$20\mu\mathrm{C}$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance