A capacitor is constructed from two metal sheets placed 0.18 mm apart. The total
ID: 1432814 • Letter: A
Question
A capacitor is constructed from two metal sheets placed 0.18 mm apart. The total capacitance is 9 pF (9×10-12 F). A battery is used to charge the capacitor to 0.15 nC (0.15×10-9 C). The positive and negative plates are shown in the figure:
8) What is the direction of the electric field lines between the plates of the capacitor? a) left b) right c) the electric field is zero in the capacitor
9) What is the area of each plate that was needed to construct this capacitor? a) 0.00018 m2 b) 0.00038 m2 c) 0.00058 m2
10) What is the strength of the electric field between the plates? a) 32,600 V/m b) 62,600 V/m c) 92,600 V/m
11) While still connected to the battery, the plates are pulled farther apart. What happens to the energy stored in the capacitor? a) the energy stored increases b) the energy stored decreases c) the energy stored remains the same
12) If 2 of these capacitors are connected in series and then this combination is connected in series with 2 more of the same capacitors that are connected in parallel, what is the net capacitance of the combination of 4 capacitors? a) 2.25 pF b) 3.6 pF c) 9 pF d) 18 pF e) 36 pF
Explanation / Answer
(8) No Figure is uploaded,
(9)
C = 9* 10^-12 F
d = 0.18 * 10^-3 m
C = A*eo/d
9* 10^-12 = (A * 8.85*10^-12)/(0.18 * 10^-3)
A = 0.000182 m^2
Correct Option - (a)
(10)
E = /eo
where, = q/A
E = q/(A*eo)
E = (0.15 * 10^-9)/(0.00018 * 8.85 * 10^-12) N/C
E = 92,600 V/m
Correct Option - (c)
(11)
U = 1/2 * CV^2
U = 1/2 * (A*eo)/d *V^2
As Plates are pulled apart, d increases So Capacitance decreases while it is connected to battery V remains same, So
The energy stored decreases.
Correct option - (b)
(12)
Ceq of 2 capacitors connected in series = (9*9)/(9+9) = 4.5 pF
Ceq of 2 capacitors connected in Parallel = 9 + 9 = 19 pF
The combination is connected in series , Ceq = (19*4.5)/(19+4.5) = 3.6 pF
Correct option - (b)
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