Respuesta :
20. How much charge is on sphere B after A and B touch and are separated?
Answer:
[tex]\boxed{q_{B}=+2q}[/tex]
Explanation:
We'll solve this problem by using the concept of electric potential or simply called potential [tex]V[/tex], which is the energy per unit charge, so the potential [tex]V[/tex] at any point in an electric field with a test charge [tex]q_{0}[/tex] at that point is:
[tex]V=\frac{U}{q_{0}}[/tex]
The potential [tex]V[/tex] due to a single point charge q is:
[tex]V=k\frac{q}{r}[/tex]
Where [tex]k[/tex] is an electric constant, [tex]q[/tex] is value of point charge and [tex]r[/tex] is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius [tex]r[/tex]. Before the sphere A and B touches we have:
[tex]V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r[/tex]
When they touches each other the potential is the same, so:
[tex]V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}[/tex]
From the principle of conservation of charge the algebraic sum of all the electric charges in any closed system is constant. So:
[tex]q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}[/tex]
Therefore:
[tex](1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}[/tex]
So after A and B touch and are separated the charge on sphere B is:
[tex]\boxed{q_{B}=+2q}[/tex]
21. How much charge ends up on sphere C?
Answer:
[tex]\boxed{q_{C}=+1.5q}[/tex]
Explanation:
First: A and B touches and are separated, so the charges are:
[tex]q_{A}=q_{B}=+2q[/tex]
Second: C is then touched to sphere A and separated from it.
Third: C is to sphere B and separated from it
So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:
Here [tex]q_{A}=+2q[/tex] and C carries no net charge or [tex]q_{C}=0[/tex]. Also, [tex]r_{A}=r_{C}=r[/tex]
[tex]V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}[/tex]
Applying the same concept as the previous problem when sphere touches we have:
[tex]k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}[/tex]
For the principle of conservation of charge:
[tex]q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q[/tex]
Finally, from the third step:
Here [tex]q_{B}=+2q \ and \ q_{C}=+q[/tex]. Also, [tex]r_{B}=r_{C}=r[/tex]
[tex]V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}[/tex]
When sphere touches we have:
[tex]k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}[/tex]
For the principle of conservation of charge:
[tex]q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q[/tex]
So the charge that ends up on sphere C is:
[tex]q_{C}=+1.5q[/tex]
22. What is the total charge on the three spheres before they are allowed to touch each other.
Answer:
[tex]+4q[/tex]
Explanation:
Before they are allowed to touch each other we have that:
[tex]q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0[/tex]
Therefore, for the principle of conservation of charge the algebraic sum of all the electric charges in any closed system is constant, then this can be expressed as:
[tex]q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q[/tex]
Lastly, the total charge on the three spheres before they are allowed to touch each other is:
[tex]+4q[/tex]
