1. A 3.58-microF capacitor is charged by being connected across a 12.0-V battery
ID: 2270030 • Letter: 1
Question
1. A 3.58-microF capacitor is charged by being connected across a 12.0-V battery.(a) Find the charge on the capacitor. (b) Find the potential energy of the charged capacitor.
*Correct answer for (a) will be in range of 20.0-70.0microC and for (b) 0.100-0.400mJ
2. A parallel-plate capacitor has an area of 2.74 cm^2 and the plates are separated by 2.08 mm with air between them. (a) Find the capacitance of this capacitor. (b) How much charge does this capacitor store when connected to a 6.00-V battery?(c) Find the magnitude of the electric eld between the two plates.
*Correct answer for (a) will be between 1.00-2.00pF, for (b) 5.00-9.90pC, and (c) 2000-5000 V/m.
Explanation / Answer
1)
a)
charge on a capacitor is given by:
Q = C*V
where C =capacitance = 3.58 uF = 3.58*10^-6 F
V = Voltage across it = 12 V
So, Q = 3.58*10^-6*12 = 4.296*10^-5 C = 42.96 uF <-----------answer
b)
Potential energy stored = 0.5*C*V^2 = 0.5*3.58*10^-6*12^2
= 2.578*10^-4 J = 0.2578 mJ <-----------answer
2)
a)
Capacitance is given by:
C = eA/d
where e = permittivity of air = 8.854*10^-12
A = area of plates = 2.74 cm2 = 2.74*10^-4 m2
d = distance between the plates = 2.08 mm = 2.08*10^-3 m
So, C = 8.854*10^-12*2.74*10^-4/(2.08*10^-3)
= 1.17*10^-12 F = 1.17 pF<-------answer
b)
Charge is given by:
Q = C*V = 1.17*10^-12*6
= 6.998*10^-12 C = 6.998 pC<----------answer
where V = 6 V
c)
Magnitude of Electric field is given by:
E = V/d
where V = 6 V
d = distance between the plates = 2.08*10^-3 m
So, E = 6/(2.08*10^-3) = 2.8846*10^3 V/m = 2884.6 V/m <---------answer
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.