Question

A 50-pF capacitor is immersed in silicone oil (its dielectric constant k = 2.6). When the capacitor is connected to a 24-V battery, (a) what will be the charge on the capacitor? (b) How much energy is stored in the capacitor?                         (Ans: a = 3.1 x 10^-9 C; b = 3.7 x 10^-8 J) please explain

Answer 1

Q = CV

      = 50*10^-12*2.6*24

       = 3.1*10^-9 C

Energy stored =

Answer 2

a)

Capacitance with dielectric

C_d=KC=2.6*50 =130 pF

Charge

Q=C*V =(130*10^-12)*24

Q=3.1*10^-9 C

b)

Energy stored

E=(1/2)C_dV²=(1/2)(130*10^-12)*24²

E=3.7*10^-8 J

Answer 3

From the realtion Q=CV

here the the capacitance has been increased from C to k*C, by immersing the capacitor in to silicone liquid with k as dilectric constant

so the charge becomes Q= k*C*V

q= 2.6*50*10^-12*24v= 3.12*10^-9 C

now the energy stored in the capacitor is U=1/2*Q*V

          U= 1/2*3.12*10^-9*24=37.44*10^-9 J

               = 3.7*10^-8 J

Answer 4

C1 =50pf

C2 =kC1 = 2.6x50 pf =130pf

a) Q =C2V = 3.12nC

b) energy stored = 0.5C2V² = 3.744x10^-8J

Answer 5

a)

given

C = 50 * 10^-12 F

V = 24 v

constant k = 2.6

then by using

Q = C * V * k

Q = 50 * 10^-12 * 24 * 2.6

Q = 1.2 * 10^-9 * 2.6

Q = 3.12 * 10^-9 c

b)

energy stored :-

by using

E = C*V^2 / 2

E = (50 * 10^-12 * (24)^2 ) / 2

E = 288800 / 2

E = 14400 * 10^-12 J

then by there is dielectric constant

E = 14400 * 10^-12 * 2.6

E = 37440 * 10^-12

E = 3.7 * 10^-8 J