Question

You are given a circuit with a switch, resistor, capacitor and battery all in series. The capacitor is initially discharged and begins charging at t=0.0000s, when the switch is closed. The capacitance is C=0.0100F, the resistance is R=80.0000Ohms, and the battery EMF-15.0000 Volts. At what time is the voltage across the capacitor half the maximum? (in seconds) Your Answer: Answer

Answer 1

1.

Voltage in charging RC circuit is given by:

V = V0*(1 - exp (-t/RC))

V0 = max Voltage

V = final Voltage across capacitor = V0/2

R = 80.0000 ohm

C = Capacitance = 0.0100 F

So,

V0/2 = V0*(1 - exp(-t/(0.0100*80.0000))

1/2 = 1 - exp(-t/0.8)

exp(-t/0.8) = 1/2

ln (exp(-t/0.8)) = ln (1/2)

-t/0.8 = -ln 2

t = 0.8*ln 2

t = 0.5545 sec = time when Voltage across capacitor will be half of max voltage

2.

Magnetic force on a moving charge in magnetic field is given by:

F = q*VxB

Direction of force on charged particle is given by right hand rule:

According to this if we point our thumb towards the direction of motion and fingers towards the direction of magnetic field than direction of palm will be direction of force on a positive charged particle and direction of back side of palm will direction of force on the negative charge.

Now in given case if we point our thumb in the direction of motion ("1", which is towards the top of screen) and fingers in the direction of magnetic field, which is in right direction, then our palm will be into the screen away from you, So Direction of force on positive charge particle will be into the screen away from you.

So Direction of magnetic force on charge = ''2''

Let me know if you've any query.

toward right = i, toward top of screen = j, out of the screen toward you = k

Mathematically: V = V j (toward top of screen)

magnetic field, B = B i (toward right)

So, F = q*(V j)x(B i) = q*V*B*(jxi) = q*V*B (-k)

-k = Into the screen away from you.

Let me know if you've any query.