A system of ( 2 ) capacitors of capacitance ( 2 F ) and ( 4 F ) is connected in series across a potential difference of ( 6 ~V ). The electric charge and energy stored in the system are

Step-by-step solution

The effective capacitance of system of two given capacitors is [ array l = 1 2 F + 1 4 F = (2+1) 4 F = 3 4 C= 4 3 F array ] Now, using ( Q=C V ) [ Q= 4 3 10⁻⁶ ~F 6 ~V =8 10⁻⁶ C ] Therefore, electric charge ( =8 C ) Now, energy stored ( = 1 2 C V²= 1 2 4 3 10⁻⁶ 6 6 ) ( =2 10⁻⁶ 2 6=24 10⁻⁶ ~J ) Therefore, energy stored ( =24 J )