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Consider the two point particles separated by a distance 'd' have charges Q1 and Q2 respectively. Particle Q2 experiences an electrostatic force of 20 mN due to particle Q1. If the charges of both particles are doubled and distance between them is also doubled, what is the magnitude of the electrostatic force between them?
- a)10 mN
- b)40 mN
- c)30 mN
- d)20 mN
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Consider the two point particles separated by a distance d have charge...
Given:
Charges of particles Q1 and Q2: Q1 and Q2
Distance between the particles: d
Force experienced by Q2 due to Q1: 20 mN
To find:
Magnitude of the electrostatic force between the particles when charges and distance are doubled.
Solution:
Let's first analyze the relationship between the electrostatic force and the charges and distance.
Coulomb's Law states that the electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically, Coulomb's Law can be represented as:
F = k * (Q1 * Q2) / d^2
Where:
F is the electrostatic force
k is the electrostatic constant
Now, let's consider the given scenario and calculate the new electrostatic force.
When the charges of both particles are doubled:
New charge of Q1 = 2 * Q1
New charge of Q2 = 2 * Q2
When the distance between the particles is doubled:
New distance = 2 * d
Substituting the new values into Coulomb's Law:
New force = k * ((2 * Q1) * (2 * Q2)) / (2 * d)^2
= k * (4 * Q1 * Q2) / (4 * d^2)
= (k * Q1 * Q2) / d^2
Since k, Q1, Q2, and d^2 are all constants, the new force is equal to the old force.
Therefore, the magnitude of the electrostatic force between the particles remains the same, which is 20 mN.
Hence, the correct answer is option D) 20 mN.
Charges of particles Q1 and Q2: Q1 and Q2
Distance between the particles: d
Force experienced by Q2 due to Q1: 20 mN
To find:
Magnitude of the electrostatic force between the particles when charges and distance are doubled.
Solution:
Let's first analyze the relationship between the electrostatic force and the charges and distance.
Coulomb's Law states that the electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically, Coulomb's Law can be represented as:
F = k * (Q1 * Q2) / d^2
Where:
F is the electrostatic force
k is the electrostatic constant
Now, let's consider the given scenario and calculate the new electrostatic force.
When the charges of both particles are doubled:
New charge of Q1 = 2 * Q1
New charge of Q2 = 2 * Q2
When the distance between the particles is doubled:
New distance = 2 * d
Substituting the new values into Coulomb's Law:
New force = k * ((2 * Q1) * (2 * Q2)) / (2 * d)^2
= k * (4 * Q1 * Q2) / (4 * d^2)
= (k * Q1 * Q2) / d^2
Since k, Q1, Q2, and d^2 are all constants, the new force is equal to the old force.
Therefore, the magnitude of the electrostatic force between the particles remains the same, which is 20 mN.
Hence, the correct answer is option D) 20 mN.
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Community Answer
Consider the two point particles separated by a distance d have charge...
Concept:
- Coulomb’s law: When two charge particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to multiplication of charges of two particles and inversely proportional to square of distance between them.
Force (F) ∝ q1 × q2
Where K is a constant = 9 × 109 Nm2/C2
Case 1:
Given that charges are: Q1 and Q00
Distance = d
Force (F1) = 20 mN
Case 2:
The charges of both particles are doubled and distance between them is also doubled.
Given that charges are: 2Q1 and 2Q2
Distance = 2d
Therefore, the force remains same and it is equal to 20 mN
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Consider the two point particles separated by a distance d have charges Q1 and Q2 respectively. Particle Q2 experiences an electrostatic force of 20 mN due to particle Q1. If the charges of both particles are doubled and distance between them is also doubled, what is the magnitude of the electrostatic force between them?a)10 mNb)40 mNc)30 mNd)20 mNCorrect answer is option 'D'. Can you explain this answer?
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