Three parallel plate capacitors C₁$C_{1}$, C₂$C_{2}$ and C₃$C_{3}$ each of capacitance 5$5\ \mu\text{F}$ are connected as shown in figure. The effective capacitance between points A and B, when the space between the parallel plates of C₁$C_{1}$ capacitor is filled with a dielectric medium having dielectric constant of 4, is:

Related Formula

• Uniform Electric Field between parallel plates:

• Potential difference between two points along the direction of the field:

Core Logic

The separation between the plates is d = 10 cm$d = 10 \ \mathrm{cm}$. The electric field E$\vec{E}$ between them is uniform and runs perpendicularly between the plates (along the horizontal direction, +x$+x$):

From the geometry shown in the figure:

• AC = 3 cm$AC = 3 \ \mathrm{cm}$ is perpendicular to E$\vec{E}$ (vertical direction).
• CB = 4 cm$CB = 4 \ \mathrm{cm}$ is parallel to E$\vec{E}$ (horizontal direction).

Step 1: Calculate potential difference between A and B

Since line AC$AC$ is perpendicular to E$\vec{E}$:

Thus, the potential difference between A$A$ and B$B$ is completely due to the horizontal displacement CB$CB$:

Substitute the values:

Thus, the potential difference between A$A$ and B$B$ is (2)/(5)V$\frac{2}{5}V$.

Pattern Recognition

Sees: Uniform electric field, potential difference over diagonal paths. Trap: Projecting along the hypotenuse length (5 cm$5\text{ cm}$) directly without considering the electric field's physical direction. Shortcut: Electric field is purely horizontal. Hence, only horizontal displacement matters. The horizontal displacement is 4 cm$4\text{ cm}$ out of 10 cm$10\text{ cm}$ total plate gap, so the potential difference is (4)/(10)V = (2)/(5)V$\frac{4}{10}V = \frac{2}{5}V$.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Related Formula

• Electric field intensity on the axis of a ring of radius R$R$ at axial distance z$z$:

• Condition for maximum electric field on the axis of a ring:

Core Logic

To find where the axial field is maximized, take the first derivative of E$E$ with respect to z$z$ and set it to zero:

We are given:

• Radius of the loop R = a√(2)$R = a\sqrt{2}$

Step 1: Calculate the peak coordinate

Substitute the radius R = a√(2)$R = a\sqrt{2}$ into the maximized coordinate condition:

Thus, the electric field reaches its peak intensity at z = a$z = a$.

Pattern Recognition

Sees: Maximum axial field location of circular charged ring. Trap: Simply picking radius R$R$ directly as the distance value, or omitting the 1/√(2)$1/\sqrt{2}$ scaling factor. Shortcut: The peak of the axial field distribution of any ring always lies at z = R√(2)$z = \frac{R}{\sqrt{2}}$. With R = a√(2)$R = a\sqrt{2}$, z$z$ resolves simply to a√(2)√(2) = a$\frac{a\sqrt{2}}{\sqrt{2}} = a$.

Chapter Mix

Class 12 Physics: Electric Charges and Fields

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Core Logic

In equilibrium, three forces act on the suspended charged bob:

• Tension T$T$ directed along the string at θ = 45°$\theta = 45^{\circ}$ with the vertical sheet.
• Weight mg$mg$ directed vertically downwards.
• Electrostatic repulsion force Fₑ = qE$F_e = qE$ acting horizontally away from the sheet.

From the balance of forces in vertical and horizontal directions:

Dividing the two equations:

Substitute the expression for E$E$:

Now plug in the given numerical values:

• m = 100~mg = 100 × 10⁻⁶~kg = 10⁻⁴~kg$m = 100\mathrm{~mg} = 100 \times 10^{-6}\mathrm{~kg} = 10^{-4}\mathrm{~kg}$
• q = +10~μ C = 10 × 10⁻⁶~C = 10⁻⁵~C$q = +10\mathrm{~\mu C} = 10 \times 10^{-6}\mathrm{~C} = 10^{-5}\mathrm{~C}$
• g = 10~m/s²$g = 10\mathrm{~m/s^2}$
• ε₀ = 8.85 × 10⁻¹²~F/m$\epsilon_0 = 8.85 \times 10^{-12}\mathrm{~F/m}$

Step 1: Final Conclusion

The charge density of the sheet is 1.77~nC/m²$1.77\mathrm{~nC/m}^2$.

Pattern Recognition

For a charge hanging near a vertical charged sheet, the equilibrium angle is governed by θ = (Fₑ)/(mg)$\tan\theta = \frac{F_e}{mg}$. For θ = 45°$\theta = 45^{\circ}$, the horizontal force equals the vertical force (Fₑ = mg$F_e = mg$). Be careful to use the field of a non-conducting sheet: E = (σ)/(2ε₀)$E = \frac{\sigma}{2\epsilon_0}$.

Chapter Mix

Class 12 Physics: Electrostatics

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Core Logic

For parallel conducting plates of large area, charges redistribute on the outer and inner faces to maintain electrostatic equilibrium. Total charge per unit area on Plate 1: q₁ = σ$q_1 = \sigma$ Total charge per unit area on Plate 2: q₂ = -2σ$q_2 = -2\sigma$

The outer surface charge density on the far sides of both plates must be equal:

Now, compute the charges on the inner facing surfaces:

• Inner face of Plate 1:

• Inner face of Plate 2:

In the region between the plates, both inner surfaces create an electric field in the same direction (away from the positive plate 1 and towards negative plate 2):

Thus, the electrostatic force on +q$+q$ is:

Step 1: Final Conclusion

The force experienced by the point charge +q$+q$ is (3σ q)/(2ε₀)$\frac{3\sigma q}{2\epsilon_0}$.

Pattern Recognition

For conducting plates with total charges Q₁$Q_1$ and Q₂$Q_2$, always calculate the outer charge first: Qouter = (Q₁+Q₂)/(2)$Q_{\text{outer}} = \frac{Q_1+Q_2}{2}$. The field inside the gap is exclusively due to the inner surfaces: E = σᵢₙₙₑᵣε₀$E = \frac{\sigma_{\text{inner}}}{\epsilon_0}$.

Chapter Mix

Class 12 Physics: Electrostatics

Related Formula

Core Logic

Let the null point where the electric field is zero be at (x, 0, 0)$(x, 0, 0)$ where 0 < x < d$0 < x < d$.

At this point, the fields due to both charges are equal in magnitude and opposite in direction:

Taking the square root on both sides:

Thus, the coordinates of the null point are ((d)/(4), 0, 0)$\left(\frac{d}{4}, 0, 0\right)$.

Step 1: Final Conclusion

The point where the electric field vanishes is ((d)/(4), 0, 0)$\left(\frac{d}{4}, 0, 0\right)$.

Pattern Recognition

For two like charges, the zero-field null point always lies along the line joining them and is closer to the smaller charge. Use the standard shortcut: x = d1 + √(q₂/q₁)$x = \frac{d}{1 + \sqrt{q_2/q_1}}$ measured from charge q₁$q_1$.

Chapter Mix

Class 12 Physics: Electrostatics