Solution:

To make the oil drop move upward with the same terminal velocity it had while moving downward, the electric force upward must balance the gravitational force downward.

Solution Steps:

1. Given Data:
- Mass of the drop, \( m = 1.6 \times 10^{-12} \, \text{g} = 1.6 \times 10^{-15} \, \text{kg} \)
- Charge on the drop, \( q = 6 \times e = 6 \times 1.6 \times 10^{-19} \, \text{C} = 9.6 \times 10^{-19} \, \text{C} \)
- Gravitational force, \( F_{\text{gravity}} = mg \), where \( g = 9.8 \, \text{m/s}^2 \)

2. Gravitational Force:
\[
F_{\text{gravity}} = (1.6 \times 10^{-15}) \times 9.8 = 1.568 \times 10^{-14} \, \text{N}
\]

3. Electric Field Required:
- To counteract this gravitational force, the electric force \( F_{\text{electric}} = qE \) must equal \( F_{\text{gravity}} \).
\[
qE = F_{\text{gravity}}
\]
\[
E = \frac{F_{\text{gravity}}}{q} = \frac{1.568 \times 10^{-14}}{9.6 \times 10^{-19}} = 3.3 \times 10^4 \, \text{N/C}
\]

Conclusion:
The required electric field magnitude is \( 3.3 \times 10^4 \, \text{N/C} \).

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Solution:

From Newton's Third Law of Motion forces acting on both the charges are equal.

\[ tan \theta = \frac{F}{mg} \]

So angle made by them with vertical will also be equal. But the value of θ will increase because force is increasing due to increase in charge.

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Solution:

To find \(\sum \overrightarrow{F_i}\), the vector sum of forces acting on each charge in the system, we can apply Newton's third law:

1. Newton's Third Law:
- For any pair of charges, the force exerted by one charge on another is equal in magnitude and opposite in direction to the force exerted by the second charge on the first.

2. Net Force on System:
- Since each pair of charges in the system exerts equal and opposite forces on each other, all internal forces cancel out in pairs.

3. Resultant Force:
- As a result, the net force on the entire system of charges is zero.

Conclusion:

The total force \(\sum \overrightarrow{F_i} = 0\).

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Solution:

To analyze the motion of the positive charge \( +q \) released from point \( (2a, 0) \), let's consider the forces acting on it due to the two fixed negative charges \( -q \) at points \( (0, a) \) and \( (0, -a) \).

Solution Steps:

1. Symmetry of Forces:
- Each negative charge \( -q \) exerts an attractive force on \( +q \) along the line joining them.
- Due to symmetry, the vertical components of these forces (along the y-axis) will cancel out, while the horizontal components (along the x-axis) will add up, pulling \( +q \) toward the y-axis.

2. Resultant Force Direction:
- The resultant force always points toward the y-axis but varies with distance from it.
- The force does not follow a linear restoring force law (i.e., it’s not proportional to displacement), which is required for simple harmonic motion (SHM).

3. Motion of \( +q \):
- The charge \( +q \) will oscillate back and forth across the y-axis due to the attractive forces from the two fixed charges.
- However, since the force is not proportional to displacement, the motion is **not SHM**.

Conclusion:
The positive charge \( +q \) will oscillate but will not execute SHM.

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Solution:

The force between two charges varies inversely with the square of the distance between them, according to Coulomb's law:

\[
F \propto \frac{1}{r^2}
\]

Solution Steps:

1. Initial and Final Distances:
- Initial distance \( r_1 = 0.06 \, \text{cm} \)
- Final distance \( r_2 = 0.06 - 0.01 = 0.05 \, \text{cm} \)

2. Force Ratio:
\[
\frac{F_2}{F_1} = \frac{r_1^2}{r_2^2}
\]

3. Substitute Values:
\[
\frac{F_2}{5} = \frac{(0.06)^2}{(0.05)^2}
\]
\[
F_2 = 5 \cdot \frac{0.0036}{0.0025} = 5 \cdot 1.44 = 7.2 \, \text{N}
\]

Conclusion:
The new force \( F_2 \) is \( 7.2 \, \text{N} \).

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Solution:

To find the dimensions of \(\frac{Kq^{2}}{Gm^{2}}\), we analyze the dimensions of each component.

1. Dimensions of \( K \):
- Given \( K = \frac{1}{4\pi \varepsilon_0} \), where \(\varepsilon_0\) is the permittivity of free space.
- \( K \) has dimensions of \(\left[ \text{Force} \cdot \text{Distance}^2 \cdot \text{Charge}^{-2} \right] = \left[ M^1 L^3 T^{-4} A^{-2} \right] \).

2. Dimensions of \( q^2 \):
- The charge \( q \) has dimensions \(\left[ A T \right]\), so \( q^2 \) has dimensions \(\left[ A^2 T^2 \right]\).

3. Dimensions of \( G \):
- \( G \) is the gravitational constant with dimensions \(\left[ M^{-1} L^3 T^{-2} \right]\).

4. Dimensions of \( m^2 \):
- The mass \( m \) has dimensions \(\left[ M \right]\), so \( m^2 \) has dimensions \(\left[ M^2 \right]\).

5. Combine Everything:
- Now, \(\frac{Kq^2}{Gm^2}\) has dimensions:
\[
\frac{\left[ M^1 L^3 T^{-4} A^{-2} \right] \cdot \left[ A^2 T^2 \right]}{\left[ M^{-1} L^3 T^{-2} \right] \cdot \left[ M^2 \right]}
\]

6. Simplify:
- This simplifies to \(\left[ M^0 L^0 T^0 A^0 \right] = \text{No Dimensions}\).

Conclusion:
The expression \(\frac{Kq^2}{Gm^2}\) is dimensionless.

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Solution:

To maximize the force between two charges \( q_1 \) and \( q_2 \) obtained by dividing a charge \( Q \) into two parts, we can use Coulomb's law:

\[
F = k \frac{q_1 q_2}{r^2}
\]

where \( r \) is the distance between them.

Steps:

1. Set up the variables:
Let \( q_1 = x \) and \( q_2 = Q - x \).

2. Express the force:
Substitute \( q_1 \) and \( q_2 \) into the formula:
\[
F = k \frac{x (Q - x)}{r^2}
\]

3. Maximize \( F \):
To find the maximum force, take the derivative of \( F \) with respect to \( x \) and set it to zero:
\[
\frac{dF}{dx} = k \frac{Q - 2x}{r^2} = 0
\]

Solving \( Q - 2x = 0 \) gives \( x = \frac{Q}{2} \).

4. Conclusion:
The force is maximum when each part is \( \frac{Q}{2} \).

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Solution:

To determine the direction of the resultant force on the positive charge \(q_0\) placed at the center due to the four charges on the clock face, we analyze the forces exerted by each charge.

Solution Steps:

1. Forces by Opposing Pairs:
- The positive charges at 12 and 3 exert a repulsive force on \(q_0\), while the negative charges at 6 and 9 exert an attractive force on \(q_0\).
- Each pair of opposing charges creates forces with equal magnitudes but different directions.

2. Force Components:
- The force due to the charges at 12 and 6 (along the vertical line) will cancel each other out in the vertical direction, as they are equal and opposite.
- Similarly, the forces due to the charges at 3 and 9 (along the horizontal line) will cancel each other out in the horizontal direction.

3. Resultant Force:
- Since the charges at 12 and 9 are positive and repulsive, they push \(q_0\) away from them.
- The vector sum of these forces results in a net force pointing diagonally between the 6 and 9 positions.

4. Direction:
- The resultant force points toward the 7:30 position, as this is the direction of the net vector resulting from the combination of all four forces.

Thus, the direction of the resultant force on \(q_0\) points to 7:30.

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Solution:

To find the time period of oscillation of the third charge \(-q\) when displaced perpendicular to the line joining the two charges of magnitude \(+Q\), we can follow these steps:

1. Restoring Force: When the charge \(-q\) is displaced by a small distance \(y\) perpendicular to the line joining the two charges, the net force on it due to the two fixed charges \(+Q\) has a restoring nature and acts toward the equilibrium position.

2. Approximation: For small \(y\), the restoring force \(F\) is proportional to \(y\), making it a simple harmonic motion (SHM) problem.

3. Electric Field: The electric field at the midpoint due to each charge \(+Q\) is \(E = \frac{Q}{4\pi\epsilon_0 (a^2 + y^2)}\). Using \(y \ll a\), we approximate the force on \(-q\) by expanding for small \(y\), resulting in a force \(F = -k y\), where \(k = \frac{2Qq}{4\pi \epsilon_0 a^3}\).

4. Angular Frequency: For SHM, the angular frequency \(\omega\) is \(\sqrt{\frac{k}{m}} = \sqrt{\frac{2Qq}{4\pi \epsilon_0 a^3 m}}\).

5. Time Period: The time period \(T\) is given by:
\[
T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{2ma^3 \pi \epsilon_0}{Qq}}
\]

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