Two charged spheres are kept at a great distance from each other. Sphere A has charge +20q and sphere B has charge -14q. A neutral conducting sphere, sphere X, is then touched upon the sphere A, then sphere B, and X, repectively, after these q(a),q(b) and q(x) are the charges on the spheres. After these contacts, what is the ordered triple [q(a)/q , q(b)/q, q(c)/q]? Don't ignore signs!
To solve this problem, we need to understand the concept of charge conservation and how charges distribute when conductors are brought into contact.
When two conductors are brought into contact, their charges redistribute until they reach a state of electrostatic equilibrium based on the principle of charge conservation. In this equilibrium, the total charge before and after the contact remains the same, but the distribution of charge may change.
Let's consider the steps in the problem one by one:
Step 1: Sphere X touches Sphere A.
- Before the contact: Sphere A has a charge of +20q, and Sphere X is neutral.
- When they touch, charges redistribute between Sphere A and Sphere X. Since Sphere X is neutral, it gains some positive charge from Sphere A. As a result, Sphere A will have less charge.
- The total charge you start with (+20q) remains the same after the contact, so the negative charge on Sphere A (-14q) must distribute to both Sphere A and Sphere X.
- The new charge on Sphere A, q(a), will be the remaining positive charge, so q(a) = +20q - (-14q) = 34q.
- The new charge on Sphere X, q(x), will be the negative charge gained, so q(x) = -14q.
Step 2: Sphere X touches Sphere B.
- Before the contact: Sphere B has a charge of -14q, and Sphere X has a positive charge of +34q (from the previous contact with Sphere A).
- When they touch, charges redistribute between Sphere B and Sphere X. Since Sphere X has a positive charge, it can share some positive charge with Sphere B.
- The total charge you start with (-14q) remains the same after the contact, so the positive charge on Sphere X (+34q) must distribute to both Sphere X and Sphere B.
- The new charge on Sphere B, q(b), will be the remaining negative charge, so q(b) = -14q - (+34q) = -48q.
- The new charge on Sphere X, q(x), will be the positive charge redistributed, so q(x) = +34q + (-34q) = 0 (neutral).
Step 3: Sphere X touches itself (no net charge change).
- After the previous contacts, Sphere X becomes neutral (q(x) = 0).
- When Sphere X touches itself, there is no charge redistribution because there's no net charge change on Sphere X.
- Therefore, q(x) remains 0.
In summary, the ordered triple [q(a)/q, q(b)/q, q(c)/q] is [34/1, -48/1, 0/1] or simply [34, -48, 0].