Given the circuit show, which has reached electrostatic equilibrium, a) Find Ce,

Question

Given the circuit show, which has reached electrostatic equilibrium,
a) Find Ce, the single capacitance equivalent to the circuit
b) Find the charge on one plate of each of the original capacitors
c) Find the change in potential (sign and magnitude) in going from point a to point b in original circuit.
Given answers: a)Ce=1.625micro f
b) q1=qe=29.2micro C , q3=20.2micro C, qs=q2=q4=9.0micro C
c)Vab =V3=-3.4V

EC-3 Given the circuit shown to the right, which has reached electrostatic equilibrium, a) Find CE, the single capacitance equivalent to the circuit. Vbat b) Find the charge on one plate of each of the original capacitors. .-18.0V C3-6 C,-8 equivalent circuit CE e) Find the change in potential (sign bar 0V Ca and magnitude) in going from point 'a' to point 'b' in the original circuit.

Explanation / Answer

Given,

r = 20 cm ; sigma = 1.5 pC/m^2

a)4 and 8 are in series

C48 = 4 x 8/(4 + 8) = 2.67 uF

Now, C48 in parallel with 6

C6-48 = 2.67 + 6 = 8.67 uF

Now, C6-48 is in series with 2

Ceq = 8.67 x 2/(8.67 + 2) = 1.63 uF

Hence, Ceq = 1.63 uF

b)Charge in the circuit is:

Q = CV = 1.63 x 10^-6 x 18 = 29.3 uC

Q1 = 29.3 uC

drop across C1 is V1 = 29.34/2 = 14.67 V

drop across parallel combination is:

Vp = 18 - 14.67 = 3.33 V

Q3 = C3V3 = 6 x 3.33 = 19.98 = 20 uC

Q2 = Q4

Q2 = Q4 = 29.2 - 20 = 9 uC

Hence, C1 = 29.34 uC ; C2 = 8.88 uC ; C3 = 20 uC ; C4 = 9 uC

c)Va - Vb = -3.33 V

(calculated in the former part)

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