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How can a charge of 3.33 X 10^-9 coulomb be distributed between two particles placed in air 5 X 10^-2 meter apart so that the repulsive force between them is a maximum ?
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How can a charge of 3.33 X 10^-9 coulomb be distributed between two pa...
Charge Distribution for Maximum Repulsive Force


When two particles are placed in air, they experience an electrostatic force of repulsion or attraction depending on their charges and distance between them. In order to find the charge distribution between two particles that maximizes their repulsive force, we can follow these steps:


Step 1: Calculate the Coulomb Force


The Coulomb force between two charged particles is given by:

F = (k * q1 * q2) / r^2

where F is the force, k is the Coulomb constant, q1 and q2 are the charges of the particles, and r is the distance between them.


Step 2: Differentiate the Force with Respect to Charge


To find the maximum repulsive force between two particles, we need to differentiate the Coulomb force with respect to the charge of one of the particles. This will give us the value of the charge that maximizes the force. By setting the derivative to zero, we get:


(dF/dq1) = (k * q2) / r^2 = 0

which implies q2 = 0


This means that one of the particles should be neutral, and the entire charge should be on the other particle.


Step 3: Calculate the Charge on the Charged Particle


Now, we need to calculate the charge on the charged particle. We can use the formula:


q1 = F * r^2 / k


where F is the maximum repulsive force we want to achieve, r is the distance between the particles, and k is the Coulomb constant.


Substituting the values, we get:


q1 = (3.33 * 10^-9) * (5 * 10^-2)^2 / (9 * 10^9) = 4.63 * 10^-10 C


So, the charge on the charged particle should be 4.63 * 10^-10 C, and the neutral particle should have no charge.
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How can a charge of 3.33 X 10^-9 coulomb be distributed between two pa...
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How can a charge of 3.33 X 10^-9 coulomb be distributed between two particles placed in air 5 X 10^-2 meter apart so that the repulsive force between them is a maximum ?
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How can a charge of 3.33 X 10^-9 coulomb be distributed between two particles placed in air 5 X 10^-2 meter apart so that the repulsive force between them is a maximum ? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about How can a charge of 3.33 X 10^-9 coulomb be distributed between two particles placed in air 5 X 10^-2 meter apart so that the repulsive force between them is a maximum ? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for How can a charge of 3.33 X 10^-9 coulomb be distributed between two particles placed in air 5 X 10^-2 meter apart so that the repulsive force between them is a maximum ?.
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