Question

A 1.00

Answer 1

1.

Q=CV =1*6=6uC

Now both capacitors are connected in parallel

equivalent capacitance

C=1+4 =5 uF

V=Q/C =6/5 =1.2 V

Q(1) =CV =1*1.2 =1.2 uF

Q(4) =4*1.2 =4.8 uC

2)

a)

Initial PE is

Ui=KqQ/r =(9*10^9)(-2.8*10^-9)(8.45*10^-9)/(1.64*10^-6)

Ui=-0.12984 J

final PE is

Uf=(9*10^9)(-2.8*10^-9)(8.45*10^-9)/(0.5*10^-6)

Uf=-0.42588 J

KE=-dPE =Ui-Uf

KE =-0.12984+0.42588

KE=0.296 J =0.3 J (approx)

b)

V=sqrt[2KE/m]

V=sqrt[2*0.3/7.7*10^-6)

V=277.3 m/s

Answer 2

charge stored in both the cap will be same if there is no leackage.

q =c*v =6 uC.

a)

as energy is conserved so the kinetic energy will be same as the W = charge*potential at that point.

so

W = 1/(4*pi*e) *(q1*q2/r)

= 0.425 J

b)

0.425 = 0.5 * 7.7*10^-6 *v^2

v = 332.593 m/S

Answer 3

After the first capacitor is charged the potential differnce across it should be =pd of the cell

So pd across 1uF cap=6 v

Now this is connected to 4.00uF,the first capacitor will charge the second one untill the potential across both the capacitor becomes equal

charge across first cap=1*10^-6*6 (Q=CV)..........(net charge)

This charge will flow across second capacitior

Q1 across 1uC

Q2 across 4uC

Q1+Q2=6*10^-6.....(1)

potential=Q/c

(6*10^-6-Q2)/1=(Q2)/4

6*10^-6/Q2-1=0.25

Q2=6*10^-6/1.25=4.8*10^-6 F

Q1=1.2*10^-6