A capacitor of capacitance `4 mu F` is charged to a potential difference of 6 V with a battery .
The battery is remvoed and in its place another capacitor of capacitance is `8 mu F` introduced and the circuit is closed. The potential difference attained by each of the capacitors in V is

Options

A. 2
B. 4
C. 6
D. 8

Answer

A

Explanation

The Final circuit of the series capacitor system is shown in figure ,

Since, we know that in a series capacitor system, the charge stored by both the capacitors is equal .
Hence, `Q= C_(eq) V`
Where , `C_(eq) = (4xx8) /(4+8) = 8/3 mu F rArr Q = 8/3 xx6 = 16 mu C`
Hence, `V_1 = Q/(C_1) = (16) /4 =4V and V_2 = Q/(C_2) =(16)/8 =2 V`
Hence the potential difference,
`Delta V = V_1 -V_2`
`rArr Delta V = 4-2 = 2V`