Two point charges 2q and q are placed at vertex A and centre of face CDEF of the cube as shown in figure. The electric flux passing through the cube is :

Related Formula

Series: Ceq = ( 1C₁ + 1C₂)⁻¹$C_{\text{eq}} = \left(\frac{1}{C_{1}} + \frac{1}{C_{2}}\right)^{-1}$ Parallel: Ceq = C₁ + C₂$C_{\text{eq}} = C_{1} + C_{2}$

Core Logic

The original capacitance is C = ε₀Ad$C = \frac{\varepsilon_{0}A}{d}$. We mentally split the system into 4 distinct block capacitors based on area and thickness splits. Upper half contains K₁$K_{1}$, acting on full Area but half depth. It can be split into two parallel halves to match the lower blocks. Left branch: Top block C₁$C_{1}$ (area A/2$A/2$, depth d/2$d/2$, K=2$K=2$) in series with Bottom block C₂$C_{2}$ (area A/2$A/2$, depth d/2$d/2$, K=3$K=3$). Right branch: Top block C₄$C_{4}$ (area A/2$A/2$, depth d/2$d/2$, K=2$K=2$) in series with Bottom block C₃$C_{3}$ (area A/2$A/2$, depth d/2$d/2$, K=5$K=5$).

Step 1: Calculate Block Capacitances

Step 2: Resolve Left and Right Series Combinations

Wait, the provided solution assigns block C₁ = 5C$C_1 = 5C$ and C₂=2C$C_2=2C$. It splits it differently: upper block Cupper = 2 ε₀ Ad/2 = 4C$C_{upper} = \frac{2 \varepsilon_{0} A}{d/2} = 4C$. Lower left CLL = 3 ε₀ A/2d/2 = 3C$C_{LL} = \frac{3 \varepsilon_{0} A/2}{d/2} = 3C$. Lower right CLR = 5 ε₀ A/2d/2 = 5C$C_{LR} = \frac{5 \varepsilon_{0} A/2}{d/2} = 5C$. Then Lower parallel CL = 3C + 5C = 8C$C_{L} = 3C + 5C = 8C$. Then Ceq = (4C × 8C)/(4C + 8C) = (32)/(12)C = (8)/(3)C$C_{eq} = \frac{4C \times 8C}{4C + 8C} = \frac{32}{12}C = \frac{8}{3}C$. Let's check the PDF solution logic: the solution calculates C₁ = 5C, C₂ = 2C$C_1 = 5C, C_2 = 2C$ in series. It seems it mismatched K values to the variable names but the math output is Ceq = (92)/(35)C ≈ 2.62C = (7.9)/(3)C n ≈ 8$C_{eq} = \frac{92}{35}C \approx 2.62C = \frac{7.9}{3}C \implies n \approx 8$. Wait, the solution logic says: Ceq = (92)/(35)C ≈ 2.62 C$C_{eq} = \frac{92}{35}C \approx 2.62 C$. The correct pure method: Upper block K₁=2$K_1=2$. C₁ = 2 ε₀ Ad/2 = 4C$C_1 = \frac{2 \varepsilon_{0} A}{d/2} = 4C$. Bottom blocks are parallel: C₂ = 3 ε₀ A/2d/2 = 3C$C_2 = \frac{3 \varepsilon_{0} A/2}{d/2} = 3C$, C₃ = 5 ε₀ A/2d/2 = 5C$C_3 = \frac{5 \varepsilon_{0} A/2}{d/2} = 5C$. Equivalent bottom = 3C + 5C = 8C$= 3C + 5C = 8C$. Total Ceq = (4C · 8C)/(4C + 8C) = (32)/(12)C = (8)/(3)C$C_{eq} = \frac{4C \cdot 8C}{4C + 8C} = \frac{32}{12}C = \frac{8}{3}C$. So exact n = 8$n = 8$. The PDF solution makes an arithmetic convolution splitting it vertically fully into two branches but correctly arrives at approximately 8. We will use the direct correct layout that yields exactly 8.

Step 3: Final Capacitance

We are given Ceq = (n)/(3) C$C_{\text{eq}} = \frac{n}{3} C$. Comparing both, we get n = 8$n = 8$.

Pattern Recognition

Sees: "composite dielectrics with splits" → Treat series divisions (depth cuts) as series capacitors and parallel divisions (area cuts) as parallel capacitors. Horizontal boundary = Series. Vertical boundary = Parallel.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Electric field due to a short dipole on its axial line:

Electric field due to a short dipole on its equatorial line:

Core Logic

Point P is at a distance of r = 40 cm = 0.4 m$r = 40 \, \mathrm{cm} = 0.4 \, \mathrm{m}$ from the centre of both dipoles. For dipole A, point P lies on the axial line. For dipole B, point P lies on the equatorial line.

Dipole moments:

Step 1: Calculate Field Components

Wait, the orientation in the figure sets the directions. From the diagram, A is oriented left-to-right (minus to plus) so E₁$\vec{E}_1$ on its axis is along + i$+\hat{i}$ or - i$-\hat{i}$ depending on precise placement. Based on standard convention:

E₂ = -(KP₂)/(r³) j$$\vec{E}_2 = -\frac{KP_2}{r^3} \hat{j}$$ (Equatorial field points opposite to dipole vector)

Net Field vector based on solution structure:

Step 2: Find Magnitude of Net Field

Wait, the solution simplifies it as:

Magnitude:

Pattern Recognition

Identify the relationship of the observation point to each dipole (axial vs equatorial). The axial field has a factor of 2, while the equatorial does not. Vector addition is mandatory since the fields are perpendicular.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Core Logic

First, derive the potential function V(r)$V(r)$ from the given external electric field.

Both charges are at distance r = 9 cm = 0.09 m$r = 9 \, \mathrm{cm} = 0.09 \, \mathrm{m}$ from the origin (where the field points radiate from).

Step 1: Calculate Energy of Individual Charges in External Field

Energy of q₁$q_1$ in the external field:

Energy of q₂$q_2$ in the external field:

Sum of self energies:

Step 2: Calculate Interaction Energy Between Charges

Distance between charges r₁₂ = 9 - (-9) = 18 cm = 0.18 m$r_{12} = 9 - (-9) = 18 \, \mathrm{cm} = 0.18 \, \mathrm{m}$.

Step 3: Total Energy

Pattern Recognition

When an external field is present, total energy is the sum of the potential energies of each charge independently in that external field, plus the mutual interaction energy between the charges themselves.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Q = CV$Q = CV$

Core Logic

When a 2 mm$2 \, \mathrm{mm}$ mica sheet is introduced, it acts as two capacitors in series: air gap (3 mm$3 \, \mathrm{mm}$) and mica gap (2 mm$2 \, \mathrm{mm}$). New charge Q₂ = 1.25 Q₁$Q_2 = 1.25 Q_1$ (since it draws 25% more). Thus, Ceq = 1.25 C$C_{eq} = 1.25 C$.

Step 1: Calculate Equivalent Capacitance

Step 2: Equate to New Charge Requirement

0.75K + 0.5 = K$0.75K + 0.5 = K$ 0.25K = 0.5$0.25K = 0.5$ K = 2$K = 2$

Pattern Recognition

When a dielectric slab of thickness t$t$ is inserted, equivalent capacitance is directly modeled as C' = (ε₀ A)/((d - t) + t/K)$C' = \frac{\epsilon_0 A}{(d - t) + t/K}$. Constant voltage implies Q ∝ C$Q \propto C$, directly yielding the ratio equation.

Chapter Mix

Class 12 Physics: Electrostatics