Question

1) The figure below shows two charged particles on one axis. The loads are free to move; however, it is possible to place a third particle at such a point that the three particles are in balance. (a) Is this point to the left of the first two particles, to the right or between them? (b) Should the charge of the third particle be positive or negative? (c) Is the balance stable or unstable? (d) What would be the electric field value at this point?
2) In the Figure below, the battery has a potential difference and the five capacitors have a capacitance of 12.0 µF. Determine the load (a) of capacitor 1; (b) of capacitor 2

Answer 1

Answer:

1(a) Between them

1(b) Negative

1(c) Unstable

1(d) 0

2(a) 12V µC

2(b) 2.4V µC

Explanation:

1) Both particles have a positive charge.  Like charges repel each other, so each particle experiences a force pushing away.

To balance the particles, there needs to be a force pushing the particles towards each other.  To do this, we must add a negative charge between them.

This balance is unstable; if we slightly nudge one of the particles, the particle will not return to its initial position.

The electric field at this point is 0.

2(a) I assume by "load" you mean the charge on the capacitor.  The voltage of the battery isn't given, so I'll leave it as V.

The charge is equal to the capacitance times the voltage.

Q = CV

Q = (12.0 µF) V

Q = 12V µC

2(b) Capacitors in series have an equivalent capacitance of:

1/C = 1/C₁ + 1/C₂

They also have the same charge.

Q = Q₁ = Q₂

Capacitors in parallel have an equivalent capacitance of:

C = C₁ + C₂

They have the same voltage.

V = V₁ = V₂

The two capacitors in series have a capacitance of:

1/C = 1/12 + 1/12

C = 6 µF

This is in parallel with another capacitor, so the equivalent capacitance is:

C = 6 + 12

C = 18 µF

This is in series with one more capacitor:

1/C = 1/18 + 1/12

C = 7.2 µF

So the total charge in the right side of the circuit is:

Q = CV

Q = (7.2 µF) V

Q = 7.2V µC

That means there's 7.2V µC on the bottom capacitor, and 7.2V µC on the combination of capacitors on the top.  So the voltage across that combination is:

Q = CV

7.2V µC = (18 µF) V

V = 0.4V

So the charge in the left branch is:

Q = CV

Q = (6 µF) (0.4V)

Q = 2.4V µC

So the charge in each capacitor in that branch is also 2.4V µC.