Electrostatic Potential and Capacitance Part 1

According to given circuit diagram,
Capacitance =50μF=50×10−6F
Supply voltage, V = 6 VIn steady state, capacitor will act as open circuit,
∴ Equivalent resistance Req=(2+2+2)kΩ=6kΩ
Circuit current, I=

V

Req

=

6

6×1000

=10−3A
∴ Voltage across 2kΩ=1×2=10−3×2×103=2V
Now, charge on capacitor, q=CV=50×10−6×2
=100×10−6C
=100μC