a charge Q is divided into two parts q and Q-q and separated by a distance R. the force of repulsion between them will be maximum when
Correct answer: B. q=Q/2
- A. q=Q/4
- B. q=Q/2
- C. q=Q
- D. none
Explanation
The force of repulsion between two charges depends on the product of their charges and the square of the distance between them, as described by Coulomb's law:F = k * (q1 * q2) / r^2where:F is the force (in Newtons)k is Coulomb's constant (approximately 8.99 x 10^9 Nm^2/C^2)q1 and q2 are the charges of the two objects (in Coulombs)r is the distance between the charges (in meters)In this scenario, we have:Q = total chargeq = one part of the divided chargeQ - q = the other part of the divided chargeR = distance between the chargesWe want to maximize the force of repulsion. Since the distance (R) and Coulomb's constant (k) are fixed in this problem, we need to consider the product of the charges (q * (Q - q))
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Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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