Three capacitors of `2muF, 3muF and 5muF` are independently charged with batteries of emf’s 5V, 20V and 10V respectively. After disconnecting from the voltage sources. These capacitors are connected as shown in figure with their positive polarity plates are connected to A and negative polarity is earthed. Now a battery of 20V and an uncharged capacitor of `4muF` capacitance are connected to the junction A as shown with a switch S. When switch is closed, find :
(a) the potential of the junction A. (b) final charges on all four capacitors.

Answer

(a) `(100)/(7)volts`;
(b) `28.56mC, 42.84mC, 71.4mC, 22.88mC`

Explanation

Charge on the capacitors
`q_(2muF) = 2muF xx 5V = 10muC`
`q_(3muF) = 3muF xx 20V = 60muC`

`q_(5muF) = 5muF xx 10V = 50muC`
Total charge on `o+` plates
`= 10muC +60muC +50muC = 120muC`
After closing the switch let `V_(A)` is the potential of point `A`.
Form conservation of charge
`q'_(2muF) +q'_(3muF) +q'_(5muF) +q_(4muF) ...(1)`
`= 120muC`
`(q_(4muF))/(4) +20 = V_(A) ...(2)`
form (1) and (2)

`V_(A) = (100)/(7)` volt
`q'_(2muF) = 2muF xx (100)/(7)V = (200)/(7) muC`
& `q'_(3muF) = 3muF xx (100)/(7)V = (300)/(7)muC`
`q'_(5muF) = 5muF xx (100)/(7) V = (500)/(7)muC`
& `q'_(4muF) = 4muF xx ((100)/(7)-20) V = -(160)/(7)muC`