Electric Charges And Fields

Step 1: State the formula for the force between two charges with an intervening medium.
The force between two point charges and separated by a distance is . When a medium of dielectric constant fills a thickness between the charges (with a total distance ), the effective distance in the denominator of the force is changed from to . The effective distance for the force calculation is given by: The force can be written as .

Step 2: Apply the given condition to find the effective distance.
The initial force in air is . The final force is . The distance remains the same. The ratio of the forces is . So, .

Step 3: Use the given thickness of the medium.
The thickness of the medium is of the total distance .

Step 4: Substitute the effective distance formula and solve for K.
We have and . Substitute these into the effective distance formula: Divide by : Squaring both sides gives the dielectric constant:

Let a particle of charge and mass be accelerated from rest in a uniform electric field .
The force on the particle is .
The acceleration of the particle is .
The particle starts from rest ( ) and covers a displacement . We can use the kinematic equation .
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We can solve this for the time taken, :
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Since the displacement and the electric field are the same for both particles, we can say that .
Let's find the mass and charge for the proton and alpha particle.
For a proton (p): Mass , Charge .
For an alpha particle ( ): Mass , Charge .
Now we can find the ratio of the times, .
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The ratio of the times taken by the proton and the alpha particle is .

Step 1: Understanding the Concept:
This problem involves Coulomb's Law and the principle of superposition. The net force on a charge is the vector sum of the individual forces exerted on it by all other charges. For the net force to be zero, the vector sum of all forces must be zero.
Step 2: Key Formula or Approach:
Coulomb's Law: The force between two point charges and separated by a distance is .
Let the charge at the origin be µC. The other charges are at cm, µC at cm, and µC at cm.
The net force on is .
Step 3: Detailed Explanation:
Let's analyze the direction of the forces on the charge µC at the origin. Let the positive x-direction be to the right.
- Force due to µC: Since and are both positive, the force is repulsive, pushing to the left (in the -x direction).
- Force due to µC: Since and are both positive, the force is also repulsive, pushing to the left (in the -x direction).
- For the net force to be zero, the force due to , which is , must balance the other two forces. This means must point to the right (in the +x direction).
- An attractive force is needed between and . Since is positive, must be negative.
Now, we can equate the magnitudes of the forces:
We can cancel and from both sides. We can also use distances in cm, as the units will cancel out in the ratio.
Since we determined that Q must be negative for the forces to balance, we have:
Step 4: Final Answer:
The value of charge Q is -1.75 µC. Therefore, option (B) is correct.

1. Let the side of the equilateral triangle be . The fourth particle of charge is placed at the midpoint of one side.
2. Distance from the two adjacent charges = , force magnitude due to each =
3. The directions of these two forces are along the lines joining the fourth particle to the adjacent vertices. They form a angle (opposite directions along the line joining the two vertices). Net force along that line = ? Wait, vector addition carefully: forces are symmetric, horizontal components cancel, vertical components add: .
4. Force from the third vertex (opposite corner): distance , force magnitude =
5. This force acts horizontally toward the midpoint. Vector sum of vertical and horizontal components gives

• The acceleration of proton: ,
acceleration of alpha particle: .
• Distance travelled under uniform acceleration: ,
with the same time for both particles.
• Therefore, the ratio of distances: • Hence, the ratio of distances travelled by proton and alpha particle in the field is 2:1.