each experiences an attractive force of 3.7 N. The spheres are then touched together and moved back to a distance r apart. Find
1 answer:
Complete Question;
Two otherwise identical, small conducting spheres have charges +6.0 µC and −2.0 µC. When placed a distance r apart,each experiences an attractive force of 3.7 N. The spheres are then touched together and moved back to a distance r apart. Find the magnitude of the new force on each sphere.
Answer:
F= 1.23 N
Explanation:
The total charge on the spheres, is + 6 μC + (-2 μC) = +4 μC.
The magnitude of the attractive force between them, according Coulomb's Law (assuming both spheres can be treated as point charges), can be expressed as follows:
(1)
Now, if we put in contact both spheres, due to the universal principle of conservation of charge, total charge on both spheres must remain the same.
Due to both spheres are conductors, this excess charge will be evenly spread out on their surfaces in contact each other, which means that each sphere, after being separated, will have a charge of +2 μC.
So, the general expression for the magnitude of the force between them, for these new values of charges, can be expressed as follows:
(2)
Dividing both sides in (1) and (2), we get:
Simplifying common terms, and rearranging, we can solve for Fx, as follows:
Fx = 3.7N / 3 = 1.23 N (the force is repulsive).
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(a) On the coil: 20 V, on the resistor: 0 V
The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:
The potential difference across the inductance is given by
(1)
where
is the time constant of the circuit
At time t=0,
So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:
(b) On the coil: 0 V, on the resistor: 20 V
Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for
Since is at the order of less than milliseconds.
Using eq.(1), we see that for , the exponential becomes zero, and therefore the potential difference across the coil is zero:
Therefore, the potential difference across the resistor will be
(c) Yes
The two voltages will be equal when:
(2)
Reminding also that the sum of the two voltages must be equal to the voltage of the battery:
And rewriting this equation,
Substituting into (2) we find
So, the two voltages will be equal when they are both equal to 10 V.
(d) at
We said that the two voltages will be equal when
Using eq.(1), and this last equation, this means
And solving the equation for t, we find the time t at which the two voltages are equal:
(e-a) -19.2 V on the coil, 19.2 V on the resistor
Here we have that the current in the circuit is
The problem says this current is stable: this means that we are in a situation in which , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law
Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.
The voltage on the coil is given by
(1)
which means that it is maximum at the moment when the battery is disconnected, when t=0:
And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.
(e-b) 0 V on both
After several seconds,
If we use this approximation into the formula
(1)
We find that
And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.