(b) Two charges of magnitude 0.610 £gC and 0.490 £gC are in a liquid dielectric with dielectric constant £e = 45.0. The charges are separated by a distance of 0.290 m.
(i) What is the size of the electric force between the charges?
(ii) By how many times is the electric force between the charges weaker in the dielectric than it would be if the particles were in a vacuum (£e = 1.00)?
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(ii) The circuit is then left for a long time with the switch closed so that the capacitor fully charges to 30.0 V. The switch is now carefully opened so that no charge leaves the capacitor and the battery is also replaced by a resistor of resistance 160. k£[.
What will be the voltage across the capacitor 160. s after the switch is closed again?
To find the answers to these questions, we will use the formulas and principles of electrostatics.
(i) The electric force between two charges is given by Coulomb's law. The formula for the electric force (F) between two charges (q1 and q2) separated by a distance (r) is:
F = k * (q1 * q2) / (r^2)
Where k is the electrostatic constant. Its value is given as:
k = 8.99 * 10^9 N m^2 C^-2
Here, we have two charges of magnitude 0.610 μC and 0.490 μC separated by a distance of 0.290 m.
Plugging in these values into the formula, we get:
F = (8.99 * 10^9 N m^2 C^-2) * (0.610 μC * 0.490 μC) / (0.290 m)^2
Calculating this, we find the size of the electric force between the charges.
(ii) To find the ratio of the electric force in the dielectric (F_dielectric) to the electric force in a vacuum (F_vacuum), we use the concept of the dielectric constant (ε).
The ratio is given by:
F_dielectric / F_vacuum = 1 / ε
Here, the dielectric constant (ε) is given as 45.0.
Plugging in this value, we find the ratio of the electric force in the dielectric to the electric force in a vacuum.
Moving on to the second part of your question:
After the switch is closed and the capacitor fully charges to 30.0 V, we can assume that the capacitor is in a steady state. This means that it is fully charged and no current is flowing through it.
When the switch is opened and the battery is replaced by a resistor, the capacitor starts discharging through the resistor. The voltage across the capacitor (Vc) at any time (t) during this discharge can be calculated using the formula:
Vc = V0 * e^(-t / RC)
Where V0 is the initial voltage across the capacitor, R is the resistance, C is the capacitance, t is the time, and e is the base of the natural logarithm. In this case, the resistance is given as 160 kΩ.
To find the voltage across the capacitor 160 s after the switch is closed again, we need to plug in the given values into the formula and calculate Vc.