A capacitor with C = 6.00 μF is fully charged by connecting it to a battery that has emf 50.0 V. The capacitor is disconnected from the battery. A resistor of resistance R = 185 Ω is connected across the capacitor, and the capacitor discharges through the resistor.
What is the charge q on the capacitor when the current in the resistor is 0.180 A?

Given the current in the resistor, the charge q on the capacitor is 1.988 × 10⁻⁴C

Given the data in the question;

First, we calculate the voltage. From Ohm's Law:

Where is the voltage, is the Current and is the resistance

We substitute in our values

Using the equation for  measuring the capacitance (C) of a capacitor:

Where q is the the charge in the capacitor, and V is the Voltage

We substitute in the equation

Therefore, Given the current in the resistor, the charge q on the capacitor is 1.988 × 10⁻⁴C

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1.99×10^-4coulombs

Explanation:

The charge (Q) across the resistor the directly proportional to the voltage (V) where capacitance of the capacitor(C) is the proportionality constant. Mathematically, Q = CV

If V is the voltage across the resistor, V = IR (according to ohm's law) where I is the current in the resistor and R is the resistance.

We need to calculate the voltage on the resistor first when 0.18A current is passed through it.

V = 0.18 × 185

V = 33.3Volts

The charge Q on the resistor will be;

Q = CV

Were C = 6.00 μF, V = 33.3

Q= 6×10^-6 ×33.3

Q = 0.0001998

Q= 1.99×10^-4Coulombs

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  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.