b.2 A pair of parallel plates with capacitance of 2 mF are connected to a 12-V b
ID: 1881281 • Letter: B
Question
b.2 A pair of parallel plates with capacitance of 2 mF are connected to a 12-V battery and separated by 20 cm (a) Find out the charge on the plate, the electric field, and the potential difference between plates. [Answer: 2.4 x 10-2 C, 60 N/, 12 VI (b) If the capacitor is disconnected from the battery and then the separation of two plates is doubled, what happens to the above quantities? 3b.3 A pair of parallel plates are connected to a 12-V battery and separated by 20 cm The area of the plate is 4.0 m2, (a) Find out the energy stored in this capacitor. IAnswer 1.27 x1* JI (b) If The capacitor is still connected with the battery and the separation of two plates is doubled, what happens to the stored energy? (c) If the capacitor is disconnected from the battery and then the separation of two plates is doubled, what happens to the stored energy? cl 3b.4 In figure, Ci-6.0 nF, C2 = 4.0 nF,V = 10 V. Calculate the charge and potential difference for each capacitor. [Answer: 2.4 10-8 C, 4 V for Ci; 2.4 x 10-8 C, 6 V for C21 3b.5 In figure, Ci = 6.0 nF, C2 = 4.0 nF, V 10 V. Calculate the charge and potential difference for each capacitor in figure (a) and (b). [Answer: 6.0 x 103 C, 10 V for Ci; 4.0 x 108 C, b) 10 V for C2lExplanation / Answer
here,
3b.2
capacitance , C = 2 mF = 0.002 F
potential difference , V = 12 V
sepration between plates , d = 20 cm = 0.2 m
a)
the charge on each plate , Q = C * V
Q = 0.002 * 12 C = 2.4 * 10^-2 C
the electric feild , E = V/d
E = 12/0.2 V/m = 60 V/m
the potential difference bbetween the plates is 12 V
b)
capacitance , C = area * e0 /d
when the plate sepration is doubled
the capacitance is halved
when the battery is disconnected
the charge remains the same
so, the potential differnce is doubled i.e 24 V
the electric feild , E = V'/d'
E = ( 2 V)/(2 d) = 60 N/C
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