Two parallel plates of infinite dimensions are uniformly charged. The surface charge density on one is \(\sigma_{A} \)and on the other is \(\sigma_{B}\), field intensity at point C will be
Net Electric Point at Point C will be directed downwards
\[ \overrightarrow{E}=\frac{\sigma_{A}}{2\varepsilon_{0}} + \frac{\sigma_{B}}{2\varepsilon_{0}}\]
Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and magnitude 27\times 10^{-22}Cm^{-2}. The electric field \overrightarrow{E} in region II in between the plates is :
Electric field at Point II will be due to both the sheets so,
\[\overrightarrow{E}=\frac{\sigma}{2\varepsilon_{0}} + \frac{\sigma}{2\varepsilon_{0}}=\frac{\sigma}{\varepsilon_{0}}\]
A point charge +20 μC is at a distance 6 cm directly above the centre of a square of side 12 cm as shown is figure. The magnitude of electric flux through the square is :
To calculate the electric flux through the square, we use Gauss's law and symmetry principles.
The total flux from a point charge \(q = +20 \, \mu\text{C}\) is:
\[
\Phi_{\text{total}} = \frac{q}{\epsilon_0}
\]
Here:
- \(q = 20 \times 10^{-6} \, \text{C}\),
- \(\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N·m}^2\).
The square is part of an imaginary cube with the charge at its center. The flux through one face of the cube (the square in question) is:
\[
\Phi_{\text{square}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0}
\]
Substitute the values:
\[
\Phi_{\text{square}} = \frac{20 \times 10^{-6}}{6 \times 8.854 \times 10^{-12}}
\]
Simplify:
\[
\Phi_{\text{square}} = 3.77 \times 10^5 \, \text{N·m}^2/\text{C}
\]
Thus, the flux through the square is approximately:
\[
{3.8 \times 10^5 \, \text{N·m}^2/\text{C}}
\]
The electric field in a region is give n by \[ \overrightarrow{E}=200\hat{i}N/C for x > 0 and -200\hat{i}N/C for x < 0 \]
A closed cylinder of length 2m and cross -section area 10² m² is kept in such away that the axis of cylinder is along X-axis and its centre coincides with origin. The total charge inside the cylinder is
(Take \(e_{0}=8.85\times 10^{-12} C^{2}m^{2}N)\)
To calculate the total charge inside the cylinder, we use Gauss's law:
\[
\Phi_E = \frac{q_{\text{inside}}}{\epsilon_0}
\]
Here:
- \(\Phi_E\) is the total electric flux through the surface of the cylinder,
- \(q_{\text{inside}}\) is the total charge enclosed by the cylinder,
- \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N·m}^2)\).
Step 1: Electric flux through the cylinder
The cylinder has two flat faces (at \(x = +1\) m and \(x = -1\) m) and a curved surface. The electric field is parallel to the axis, so the flux through the curved surface is **zero**. Thus, only the two flat faces contribute to the flux.
- For the face at \(x = +1\) m:
\(\Phi_E^+ = E \cdot A = 200 \cdot 10^{-2} = 2 \, \text{N·m}^2/\text{C}\),
- For the face at \(x = -1\) m:
\(\Phi_E^- = E \cdot A = -200 \cdot 10^{-2} = -2 \, \text{N·m}^2/\text{C}\).
The total flux is:
\[
\Phi_E = \Phi_E^+ + \Phi_E^- = 2 - (-2) = 4 \, \text{N·m}^2/\text{C}
\]
Step 2: Total charge inside the cylinder
Using Gauss's law:
\[
q_{\text{inside}} = \Phi_E \cdot \epsilon_0 = 4 \cdot (8.85 \times 10^{-12})
\]
Simplify:
\[
q_{\text{inside}} = 35.4 \times 10^{-8} \, \text{C}
\]
Thus, the total charge inside the cylinder is:
\[
{35.4 \times 10^{-8} \, \text{C}}
\]
A charged ball B hangs from a silk thread S which makes an angle θ with a large charged conducting sheet P, as shown in the figure. The surface charge density ρ of the sheet is proportional to :
To solve for the proportionality of the surface charge density \(\sigma\) with \(\tan\theta\), we analyze the forces acting on the charged ball \(B\):
Step 1: Forces acting on the ball
1. Gravitational force (\(F_g\)): Acts vertically downward, magnitude \(F_g = mg\), where \(m\) is the mass of the ball.
2. Electric force (\(F_e\)): Acts horizontally, due to the electric field produced by the charged conducting sheet.
3. Tension (\(T\)): Acts along the silk thread, balancing the net forces in both horizontal and vertical directions.
The electric field near a charged conducting sheet with surface charge density \(\sigma\) is:
\[
E = \frac{\sigma}{2\epsilon_0}
\]
The electric force on the ball is:
\[
F_e = qE = q \cdot \frac{\sigma}{2\epsilon_0}
\]
Step 2: Force balance
At equilibrium:
- In the vertical direction: \(T \cos\theta = mg\)
- In the horizontal direction: \(T \sin\theta = F_e = q \cdot \frac{\sigma}{2\epsilon_0}\)
Taking the ratio of the horizontal and vertical components:
\[
\tan\theta = \frac{T \sin\theta}{T \cos\theta} = \frac{F_e}{F_g} = \frac{q \cdot \frac{\sigma}{2\epsilon_0}}{mg}
\]
\[
\tan\theta \propto \sigma
\]
Final Answer:
The surface charge density \(\sigma\) of the sheet is proportional to:
\[
{\tan\theta}
\]
Three identical conducting plates are shown in figure. Findout the charge on the right face of the plate b.
Yes, direct formulas can simplify the solution for problems like this involving conducting plates. Here's how:
Formula for charge redistribution in conductors:
For three parallel conducting plates with charges \( Q_1, Q_2, Q_3 \):
1. The charge on the inner faces is:
\[
Q_{\text{inner}} = \frac{Q_1 - Q_3}{2}
\]
2. The charge on the outer faces remains equal to the net charge on the respective plates:
- Outer left face: \( Q_1 \)
- Outer right face: \( Q_3 \)
---
Applying the formula:
- Given charges:
\[
Q_1 = -2Q, \, Q_2 = Q, \, Q_3 = -Q
\]
- Charge on the inner faces of plate \( b \):
\[
Q_{\text{inner}} = \frac{Q_1 - Q_3}{2} = \frac{-2Q - (-Q)}{2} = \frac{-2Q + Q}{2} = \frac{-Q}{2}
\]
- Since the **right face of \( b \)** contributes to the inner faces:
\[
Q_{\text{right face of \( b \)}} = 0
\]
Final Answer:
\[
{Zero (0)}
\]