Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen. Reason R: Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell. In the light of the above statements, choose the correct answer from the options given below:

Related Formula

Core Logic

The resultant dipole moment for a system of charges is given by:

Substituting the given values:

Pattern Recognition

For a system of point charges where the net charge is non-zero, the dipole moment depends on the origin. However, taking the standard formula Σ qᵢ rᵢ$\sum q_i r_i$ yields the required mathematical expression directly.

Chapter Mix

Class 12 Physics: Electric Charges and Fields

Related Formula

Core Logic

Given potential V = ar³ + b$V = ar^3 + b$. The electric field E$E$ is given by the negative gradient of potential:

Using Gauss's Law to find the enclosed charge for a spherical region:

Step 1: Enclosed Charge Calculation

The surface area of a sphere of radius r=1$r=1$ is A = 4π(1)² = 4π$A = 4\pi(1)^2 = 4\pi$. The electric field at r=1$r=1$ is:

So,

Comparing with the given expression α × π a ε₀$\alpha \times \pi a \epsilon_0$, we get: α = -12$\alpha = -12$

Pattern Recognition

For spherical symmetry, extracting total charge enclosed is fastest using Gauss's Law at the boundary surface rather than integrating local charge density ρ(r)$\rho(r)$ using Poisson's equation.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

where reffective = (rshell, robservation)$r_{\text{effective}} = \max(r_{\text{shell}}, r_{\text{observation}})$.

Core Logic

Potential at surface of inner sphere A (radius a$a$):

Potential at surface of middle sphere B (radius b$b$): For charge q₁$q_1$, this is an outside point. For q₂$q_2$, it's on the surface. For q₃$q_3$, it's inside.

Potential at surface of outer sphere C (radius c$c$): For charges q₁$q_1$ and q₂$q_2$, this is an outside point.

Step 1: Selection of Correct Option

Matching these results shows option (3) represents the potentials perfectly.

Pattern Recognition

Potential on a sphere from inner charges uses its own radius, whereas potential from outer shells uses the outer shell's radius. Inner charges "collapse" computationally to the sphere's center.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Core Logic

Let C = (ε₀ A)/(d)$C = \frac{\epsilon_0 A}{d}$. We decompose the configurations into equivalent circuits.

For CA$C_A$:

Step 1: Calculate CB

For CB$C_B$:

Step 2: Calculate CC

For CC$C_C$:

Step 3: Comparison

Since K₁ > K₂$K_1 > K_2$: Comparing CA$C_A$ and CC$C_C$, and CC$C_C$ and CB$C_B$, algebraic manipulation proves: CA > CC > CB$C_A > C_C > C_B$

Pattern Recognition

Symmetry dictates the capacity. Adding more of the higher dielectric constant material (K₁$K_1$) in parallel paths effectively boosts total capacitance significantly, while stacking lower K₂$K_2$ diminishes it.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

where dQ = λ dx$dQ = \lambda dx$

Core Logic

Linear charge density of the wire:

Taking a small element dx$dx$ at distance x$x$ from point charge q$q$, the force is:

Step 1: Integration

Step 2: Values Substitution

Wait, computing directly from the exact equation provided:

Pattern Recognition

For a point charge q$q$ acting on a line charge of length L$L$ separated by distance a$a$, F = (k q Q)/(a(a+L))$F = \frac{k q Q}{a(a+L)}$. Directly applying this bypasses the integration completely.

Chapter Mix

Class 12 Physics: Electrostatics