In the given figure, the charge stored in 6$6\mu\text{F}$ capacitor, when points A and B are joined by a connecting wire is ________ $\mu\text{C}$.

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

Related Formula

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

Related Formula

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