Match List-I with List-II.

List-I	List-II
(A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R.	(I) σ / ε₀$\sigma / \epsilon_0$
(B) Electric field at distance r > 0 from a uniformly charged infinite plane sheet with surface charge density σ.	(II) σ / 2ε₀$\sigma / 2\epsilon_0$
(C) Electric field outside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R	(III) 0
(D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ.	(IV) σ R² / ε₀ r²$\sigma R^2 / \epsilon_0 r^2$

Choose the correct answer from the options given below:

Related Formula

Alternatively, treat as two capacitors in series: Ceq = (C₁ C₂)/(C₁ + C₂)$C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$

Core Logic

Initial capacitance (vacuum): C = A ε₀d$C = \frac{A \epsilon_{0}}{d}$.

The system can be modelled as two capacitors in series:

• A vacuum capacitor of thickness d - (d)/(3) = (2d)/(3)$d - \frac{d}{3} = \frac{2d}{3}$
• A dielectric capacitor of thickness (d)/(3)$\frac{d}{3}$ and dielectric constant K$K$.

Step 1: Capacitors in Series

The capacitances are:

Now, equivalent capacitance in series:

Pattern Recognition

Partial dielectric filling of thickness t$t$: Use formula Cnew = (ε₀ A)/(d - t + t/K)$C_{\text{new}} = \frac{\epsilon_0 A}{d - t + t/K}$. Plugging t = d/3$t = d/3$ directly gives (ε₀ A)/(d - d/3 + d/3K) = (ε₀ A)/((2d)/(3) + (d)/(3K)) = (3K ε₀ A)/(2Kd + d) = (3KC)/(2K+1)$\frac{\epsilon_0 A}{d - d/3 + d/3K} = \frac{\epsilon_0 A}{\frac{2d}{3} + \frac{d}{3K}} = \frac{3K \epsilon_0 A}{2Kd + d} = \frac{3KC}{2K+1}$.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Core Logic

Work done by external agent: Wₑₓₜ = Δ U$W_{\text{ext}} = \Delta U$, where Δ U$\Delta U$ is the change in potential energy.

Calculate the distances of points A and B from the origin: rᵢ = |A| = √(4² + 4² + 2²) = √(16+16+4) = √(36) = 6 m$r_i = |A| = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16+16+4} = \sqrt{36} = 6\text{ m}$ rf = |B| = √(2² + 2² + 1²) = √(4+4+1) = √(9) = 3 m$r_f = |B| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3\text{ m}$

Step 1: Calculate Work Done

Pattern Recognition

Electric field is conservative. Work done simply equals change in kqq/r$kqq/r$ from initial to final radial coordinate. No path dependence.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Core Logic

Using energy conservation from the initial state (separation 4R$4R$) to the final state (just touching, so center-to-center separation is 2R$2R$). Both spheres have mass m$m$ and speed u$u$.

Initial Energy:

Final Energy (just touching implies final velocity is zero):

Step 1: Equating Energies

Step 2: Final Conclusion

Pattern Recognition

When dealing with two forces (electrostatic repulsion and gravitational attraction), their potentials simply superimpose linearly. The change in total potential energy equals the loss in kinetic energy.

Chapter Mix

Class 12 Physics: Electrostatics Class 11 Physics: Gravitation

Related Formula

Q = C Vc$Q = C V_c$ where Vc$V_c$ is the steady-state voltage across the capacitor.

Core Logic

In steady state, the capacitor acts as an open circuit (blocks DC current). We must analyze the active branches.

The branch with the reversed-biased diode will carry no current. Current flows through the outer loop via the forward-biased diode. Total active resistance Req = 1 Ω + 4 Ω = 5 Ω$R_{\text{eq}} = 1\,\Omega + 4\,\Omega = 5\,\Omega$.

Current in the steady state:

Step 1: Voltage Calculation

The voltage across the capacitor Vc$V_c$ is equal to the voltage drop across the 4 Ω$4\,\Omega$ resistor, because the branch is connected in parallel.

Step 2: Final Conclusion

The charge stored is: Q = C Vc$Q = C V_c$

Pattern Recognition

Capacitor in steady-state DC = open wire. Analyze only the paths where current can physically flow, check diode polarity, find nodal voltage across the capacitor terminals, apply Q=CV.

Chapter Mix

Class 12 Physics: Electrostatics Class 12 Physics: Current Electricity Class 12 Physics: Semiconductor Electronics

Related Formula

Core Logic

By symmetry and vector addition of electric fields due to the point charges placed on the circle:

Pattern Recognition

Sees: Symmetrical charge distribution on a circle. Shortcut: Resolve components symmetrically and combine vector contributions. Check: Matches option (2). ✓

Chapter Mix

Class 12 Physics: Electrostatics