An emf source with emf of 200 V, a resistor with R =50.0 Ω, and a capacitor with C = 4.0 mF are connected in series. As the capacitor charges, when the current in the resistor is 1.00 A, what is the magnitude of the charge on each plate of the capacitor?

0.6 C

Explanation:

From the question,

Since the Resistor and the capacitor are connected in series.

The voltage across the resistor + Voltage across the capacitor = Total voltage.

Vt = Vr+Vc Equation 1

Where Vt = Total voltage, Vr = Voltage across the resistor, Vc = Voltage across the capacitor

make Vc the subject of the equation

Vc = Vt-Vr Equation 2

From ohm's law,

Vr = IR equation 3

Where I = current, R = Resistance.

Given: I = 1 A, R = 50 Ω

Substitute into equation 3

Vr = 1(50)

Vr = 50 V

Also given: Vt = 200 V

Substitute into equation 2

Vc = 200-50

Vc = 150 V.

Applying,

Q = CVc Equation 4

Where Q = charge on each plate of the capacitor, C = Capacitance of the capacitor

Given: C = 4.0 mF = 4.0×10⁻³ F, Vc = 150 V

Substitute into equation 4

Q = 4.0×10⁻³(150)

Q = 0.6 C

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s