Question

An isolated parallel-plate capacitor (not connected to a battery) has a charge of Q = 2.4×10−5C. The separation between the plates initially is d = 1.2 mm, and for this separation the capacitance is 3.1×10−11F. Calculate the work that must be done to pull the plates apart until their separation becomes 8.3 mm, if the charge on the plates remains constant. The capacitor plates are in a vacuum.

Express your answer using two significant figures

Answer 1

Solution :

Given that

charge of capacitor Q= 2.4*10^-5 C

Intial Distance d1 =1.2 mm

Seperation Capacitance C= 3.1*10^-11 F

Final Distance d2 = 8.3 mm

We have to find the work done to pull the plates untill get a final distance i.e; 8.3*10³m

• For capacitor 1

                    C1 = (e₀A)/d₁

                    C1 = ((8.85*10^-11)(A))/(1.2*10^-3)m

                    C1 = (7.37*10^-8) A

• For Capacitor 2

                    C2 = (e₀A)/d₂

                    C1 = ((8.85*10^-11)(A))/(8.3*10^-3)m

                    C1 = (1.066*10^-8) A

• C1/(7.37*10^-8) = c2/(1.066*10^-8)

            

C1/c2= (7.37*10^-8) / (1.066*10^-8)

C1/c2 = 6.98----------------(1)

• C2= C1/6.98

C2 = (3.1*10^-11) / 6.98

C2 = (0.444*10^-11)

• Q = CV

Q₁/C₁ = V₁

(2.4*10^-5)/(3.1*10^-11) = V1

V1 = 0.7741935 v

• Q2/C2=V2

(2.4*10^-5)/( 0.444*10^-11 =V2

V2= 5.4*10⁶ v

• U1= (1/2) C1V1²

U1= 0.5 C1V1²

U1= 0.5*(3.1*10^-11)( 0.7741935)²

U1= 9.29*10^-11 J

• U2= (1/2) C2V2²

U2= 0.5*(0.444*10^-11)*( 5.4*10⁶)²

U2 = 64.73 J

• Work Done = U2-U1

                    = (64.73 J)-( 9.29*10^-11 J)

                    = 64.735 J

Therefore  the work done to pull the plates untill get a final distance is 64.735 J