Two charges q₁$q_{1}$ and q₂$q_{2}$ are separated by a distance of 30 cm. A third charge q₃$q_{3}$ initially at 'C' as shown in the figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to movement of q₃$q_{3}$ from C to D is given by (q₃ K)/(4π ε₀)$$\frac{q_3 K}{4\pi \varepsilon_0}$$ , the value of K is:

Related Formula

Core Logic

Since both capacitors are connected in parallel across the same battery, their potential difference V$\mathrm{V}$ is identical.

From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first:

Using the relation q = CV$\mathrm{q} = \mathrm{C}\mathrm{V}$, since V$\mathrm{V}$ is identical, we get:

Now, the electrostatic energy stored in a capacitor is given by:

Since C₂ > C₁$\mathrm{C}_2 > \mathrm{C}_1$ and V$\mathrm{V}$ is constant, the stored energy satisfies:

Step 1: Final Conclusion

Thus, the correct relationships are C₂ > C₁$\mathrm{C}_2 > \mathrm{C}_1$ and U₂ > U₁$\mathrm{U}_2 > \mathrm{U}_1$.

Pattern Recognition

Sees parallel connection → Immediately lock potential difference V$\mathrm{V}$ as constant. This simplifies q ∝ C$\mathrm{q} \propto \mathrm{C}$ and U ∝ C$\mathrm{U} \propto \mathrm{C}$, creating a direct linear bridge from graph height to energy capacity.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Related Formula

Core Logic

The potential at a point due to a line charge on an axis is proportional to the logarithm of its perpendicular distance.

For the wire along the z-axis: Vz = -kλ ln(x² + y²)$\mathrm{V}_z = -\mathrm{k}\lambda \ln(x^2 + y^2)$

For the wire along the x-axis: Vₓ = -kλ ln(y² + z²)$\mathrm{V}_x = -\mathrm{k}\lambda \ln(y^2 + z^2)$

For the wire along the y-axis: Vy = -kλ ln(z² + x²)$\mathrm{V}_y = -\mathrm{k}\lambda \ln(z^2 + x^2)$

Summing the individual potentials to find the net configuration potential:

For an equipotential surface, set Vₙₑₜ = constant$\mathrm{V}_{\text{net}} = \text{constant}$:

Step 1: Final Expression

This maps perfectly to option (3).

Pattern Recognition

Logarithmic combination rules transform scalar potential additions into products inside the functional argument: Σ ln(rᵢ²) = ln(Π rᵢ²)$\sum \ln(\mathrm{r}_i^2) = \ln(\prod \mathrm{r}_i^2)$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Core Logic

Applying the work-energy balance for the system as the particle moves from its initial position to the x-axis:

The displacement parallel to the uniform electric field lines equals ( )/(2)$\frac{\ell}{2}$. Therefore, computing the work done by the electrostatic field:

Step 1: Final Expression

This matches option (3).

Pattern Recognition

Keep your focus on displacement along the field lines. Work relies entirely on the parallel displacement component (dₓ = 60°$d_x = \ell \cos 60^{\circ}$), completely ignoring any vertical movement components.

Chapter Mix

Class 12 Physics: Electric Charges and Fields

Related Formula

Capacitance with dielectric:

Series Capacitors:

Parallel Capacitors:

Core Logic

Let C₀ = (ε₀ A)/(d)$C_0 = \frac{\varepsilon_0 A}{d}$ be the capacitance without any dielectric.

• First Configuration (Series connection):

The dielectrics split the gap vertically, so the effective thickness of each slab is d/2$d/2$, and the area remains A$A$.

Since they are in series:

• Second Configuration (Parallel connection):

The dielectrics split the area horizontally, so the effective area of each slab is A/2$A/2$, and the distance remains d$d$.

Since they are in parallel:

Step 1: Calculating the Ratio

Now, compute (C₁)/(C₂)$\frac{C_1}{C_2}$:

Pattern Recognition

For dielectric-filled capacitors: splitting the gap (d/2$d/2$) leads to a series combination, while splitting the plate area (A/2$A/2$) leads to a parallel combination. Shortcut: Cₛₑᵣᵢₑₛ = harmonic mean$C_{\text{series}} = \text{harmonic mean}$, Cparallel = arithmetic mean$C_{\text{parallel}} = \text{arithmetic mean}$. The ratio (C₁)/(C₂)$\frac{C_1}{C_2}$ is always the ratio of the harmonic mean of the dielectric constants to their arithmetic mean!

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Related Formula

For a uniformly charged spherical shell of radius R$R$ and charge Q$Q$:

• Inside and on the surface of the shell (r ≤ R$r \le R$):

• Outside the shell (r > R$r > R$):

Core Logic

Let's calculate the potentials at the specified positions:

• Given surface potential at R = 10~cm$R = 10\mathrm{~cm}$ is 120~V$120\mathrm{~V}$.
• At the center (r = 0$r = 0$):

Since the center lies inside the shell (0 < 10~cm$0 < 10\mathrm{~cm}$), the potential equals the surface potential:

• At r = 5~cm$r = 5\mathrm{~cm}$:

Since 5~cm$5\mathrm{~cm}$ is also inside the shell (5 < 10~cm$5 < 10\mathrm{~cm}$), the potential remains constant at the surface value:

• At r = 15~cm$r = 15\mathrm{~cm}$:

Since 15~cm$15\mathrm{~cm}$ is outside the shell (15 > 10~cm$15 > 10\mathrm{~cm}$), the potential decreases inversely with distance:

Therefore, the potentials are 120~V$120\mathrm{~V}$, 120~V$120\mathrm{~V}$, and 80~V$80\mathrm{~V}$ respectively.

Pattern Recognition

The electric field inside a uniformly charged conducting spherical shell is zero, meaning that no work is done moving a charge inside it. Consequently, the potential remains absolutely uniform/constant from the surface all the way to the center!

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance