Q.1 Consider the circuit shown in the Figure. The battery potential is 40 V and

Question

Q.1 Consider the circuit shown in the Figure. The battery potential is 40 V and quati 31 V-10 V C,-0.5 F 40 V (a) Find the values for Vx, Qx, and Cx (b) What is the total capacitance present in this circuit? (c) How much total energy is stored in the circuit? A capacitor is constructed from two square, metallic plates of sides L and separation, d. The capacitor is fully charged by the power supply and then disconnected. A material of dielectric constant ? is inserted a distance x into the capacitor as shown in the Figure. Q.2 30 Find the equivalent capacitance of the device Calculate the energy stored in the capacitor with the dielectric in place. Find the magnitude and direction of the force exerted by the plates on the dielectric lf x = L/2, assuming L = 10.00 cm, d = 1.00 cm, the dielectric constant K = 4.5, and the capacitor was charged to 200 V before the dielectric was inserted calculate a numerical value for the following quantities (a) (b) (c) E-d (d) Capacitance before and after the dielectric is inserted. (e) Charge Q before and after the slab is inserted. ( Calculate the potential difference between the plates with the dielectric in place. Q.3 When switch S is thrown to the left, the plates of the C1 10 uF capacitor gets

Explanation / Answer

Remember in capacitors parallel combination voltage distribution in each part will be same and in series combination charge distribution in each capacitor will be same, So Now

Q1 = C1*V1 = 10*2 = 20 uC

Since C1, C2, and C3x are in series of each other, so they will have equal charge

C3x = Combine capacitance of C3 and Cx = (C3 + Cx)

V2 = Q2/C2 = 20/6 = 3.33 V

Q3x = 20 uC

V3x = Q3x/C3x = 20/C3x

Also V3x = V - V1 - V2 = 40 - 10 - 3.33 = 26.67 V

20/C3x = 26.67

C3x = 20/26.67 = 0.75 uF

C3x = C3 + Cx = 0.5 + Cx = 0.75

Cx = 0.25 uF

Vx = V3 = 26.67 V

Qx = Cx*Vx = 0.25*26.67 = 6.67 uC

Part B

1/Ceq = 1/C1 + 1/C2 + 1/C3x

Ceq = (1/2 + 1/6 + 1/0.75)^-1 = 0.5 uF

Part C

Energy stored = U

U = 0.5*C*V^2

U = 0.5*0.5*10^-6*40^2 = 4*10^-4 J

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