Electric Charges And Fields

and

By Gauss theorem =

Centripetal force and electrostatic force Equating both equation The time period of revolution

Step 1: Understanding the Concept:
Torque on an electric dipole: .

Step 2: Detailed Explanation:




Step 3: Final Answer:
Torque = .

Concept: Electric field near conductor is always perpendicular to surface.

Step 1: Field direction.

• Field lines terminate on conductor
• Direction is normal to surface

Step 2: Conclusion.
Field is towards the plane perpendicularly.

Concept:
• Electric field:

Step 1: Charge


Step 2: Radius


Step 3: Compute
Final Conclusion:
Option (E)

Concept:
Physics - Comparison of Fundamental Forces.
Step 1: State the formulas for both forces.

• Electric Force ( ):
• Gravitational Force ( ):

Step 2: Set up the ratio.
Since both forces follow the inverse square law ( ), the distance cancels out: $ k \approx 9 \times 10^9 ^2 ^{-2} G \approx 6.67 \times 10^{-11} ^2 ^{-2} q_e = q_p \approx 1.6 \times 10^{-19} m_e \approx 9.1 \times 10^{-31} m_p \approx 1.67 \times 10^{-27} 2.27 \times 10^{39} 10^{39}$.

Step 1: Understanding the Concept:
In electrostatics, the surface of any conductor acts as an equipotential surface. This means there can be no potential difference between any two points on the surface, and consequently, no component of the electric field parallel to the surface.

Step 2: Key Formula or Approach:
The direction of the electric field is related to the potential by . Since the surface is equipotential, the gradient along the surface is zero. Therefore, the entire electric field vector must be perpendicular (normal) to the local surface.
Furthermore, electric field lines originate from positive charges and terminate on negative charges.

Step 3: Detailed Explanation:
1. Perpendicularity: Because the spherical conductor's surface is an equipotential surface, the electric field lines immediately outside it must be strictly perpendicular (normal) to the surface at every point.
2. Direction: The conductor is described as "positively charged". Electric field lines represent the direction of force on a positive test charge. A positive test charge placed outside the sphere would be repelled. Therefore, the field lines must point away from the surface.
Combining these two facts for a spherical geometry: lines perpendicular to a spherical surface point along the radius. Since they point away, the direction is defined as "radially outward".

Step 4: Final Answer:
The direction is radially outward.

Step 1: Understanding the Concept:
When smaller liquid drops or spheres coalesce (merge) into a larger one, two fundamental quantities are conserved:
1. Total charge: The charge of the big sphere is the sum of charges of small spheres.
2. Total volume: The volume of the big sphere is the sum of volumes of small spheres.
Surface charge density ( ) relates charge to surface area.

Step 2: Key Formula or Approach:
Volume of a sphere:
Surface charge density:
Use volume conservation to find the new radius, then compute the new density.

Step 3: Detailed Explanation:
Let and be the radius and charge of a small sphere.
Let and be the radius and charge of the big sphere formed from small spheres.
1. Charge conservation:
Total charge
2. Volume conservation:
Volume of big sphere = Volume of small sphere


Taking the cube root:

3. Surface charge density ratio:
For a small sphere:
For the big sphere:
Substitute and :




The ratio is .

Step 4: Final Answer:
The ratio is 4:1.

Step 1: Understanding the Concept:
For an oil drop to be in equilibrium (stationary or moving with constant velocity), the net force acting on it must be zero. The two primary vertical forces acting on the drop in this setup are the downward gravitational force and the upward electric force.

Step 2: Key Formula or Approach:
Gravitational force acting downwards:
Electric force acting upwards:
At equilibrium, these forces are equal in magnitude and opposite in direction.


Step 3: Detailed Explanation:
Equating the magnitudes of the two forces:


To find the charge , rearrange the equation:


Step 4: Final Answer:
The correct option is (A) .

Concept:
• Field emission requires very strong electric fields

Step 1: Typical value
Field emission occurs at extremely high electric fields of order: Final Conclusion:
Option (B)