Capacitors
When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the diameter of this resistor is now tripled, the current will be
A) 2/9 A.
B) 2/3 A.
C) 2 A.
D) 3 A.
E) 18 A.
E) 18 A
V = I*R
I = V/R = V*A / ρ*l = V*(pi*D24) / ρ*l
I is proportional to D2
I2 = I1*D22/D12
I2 = 2A * (32) = 18 A
If you want to be able to create larger electric fields and store more energy, would you add a polar or non polar dielectric to your capacitor?
polar
Polar dielectrics have dipole moments and positive an negative ions do not coincide. This goes back to the idea that when a charge is applied the positive and negative ions can separate an create a larger electric field.
Two cables of the same length are made of the same material, except that one cable has twice the diameter of the other cable. When the same potential difference is maintained across both cables, which of the following statements are true? (There may be more than one correct choice.)
A) The same current flows through both cables.
B) Both cables carry the same current density.
C) The electrons have the same drift velocity in both cables.
D) The current in the thin cable is twice as great as the current in the thick cable.
E) The current in the thin cable is four times as great as the current in the thick cable.
B) and C)
J = I/A ; the changes in current and cross sectional area between the two wires cancel out
vd = I / n*q*A ; the changes in current and cross sectional area between the two wires cancel out
You are given a copper bar of dimensions 3 cm × 5 cm × 8 cm and asked to attach leads to it in order to make a resistor. If you want to achieve the SMALLEST possible resistance, you should attach the leads to the opposite faces that measure
A) 3 cm × 5 cm.
B) 3 cm × 8 cm.
C) 5 cm × 8 cm.
D) Any pair of faces produces the same resistance.
C) 5 cm × 8 cm
Bigger cross sectional area --> Less resistance
An isolated 5.40 μF parallel-plate capacitor has 4.80 mC of charge. An external force changes the distance between the electrodes until the capacitance is 2.10 μF. How much work is done by the external force?
U1 = 0.5*Q2/C
U1 = 0.5*(4.8*10-3)2/(5.4*10-6) = 2.13 J
U2 = 0.5*Q2/C
U2 = 0.5*(4.8*10-3)2/(2.1*10-6) = 5.49 J
W = U2 - U1 = 3.36 J
A parallel-plate capacitor has a capacitance of 10 mF and is charged with a 20-V power supply. The power supply is then removed and a dielectric material of dielectric constant 4.0 is used to fill the space between the plates. What is the voltage now across the capacitor?
Q = CV
C is proportional to 𝛋
V is inversely proportional to 𝛋
V2 = V1(C1/C2) = 20*(10mF/4*10mF)
V2 = 5V
Calculate the current through a 10.0-m long 22 gauge (having radius 0.321 mm) nichrome wire if it is connected to a 12.0-V battery. The resistivity of nichrome is 100 × 10-8 Ω · m.
R = ρ*l / A
R = (100*10-8)*10/ (pi*(0.321*10-3)2) = 30.89
V = I*R --> I = V/R
I = 12 V / 30.89 Ω
I = 0.388 A
A wire of resistivity ρ must be replaced in a circuit by a wire of the same material but 4 times as long. If, however, the resistance of the new wire is to be the same as the resistance of the original wire, the diameter of the new wire must be _____
A) the same as the diameter of the original wire.
B) 1/2 the diameter of the original wire.
C) 1/4 the diameter of the original wire.
D) 2 times the diameter of the original wire.
E) 4 times the diameter of the original wire.
D) 2 times the diameter of the original wire.
R = ρ*l / A
What is the energy density in the electric field at the surface of a 2.30-cm-diameter sphere charged to a potential of 2000 V?
r = d/2 = 2.3/2 = 1.15
E = -dV/dx
E = -2000/(1.15*10-2) = -173913
u = 0.5*ε0*E2
u = 0.5*(8.85*10-12)(-173913)2
u = 0.1338 J/m3
Two 4.6-cm diameter metal disks are separated by a 0.21-mm thick piece of paper. (Dielectric constant for paper is 3.7, its dielectric strength is 16 × 106 V/m.)
a. What is the capacitance?
b. What is the maximum potential difference between the disks?
a. C = (ε0*𝛋*A)/d
C = (8.85*10-12)*3.7*pi(2.3*10-2)2/(0.21*10-3)
C = 2.59*10-10 F
b. V = E*d
V = (16*106)*(0.21*10-3)
V = 3360 V
If an electric device delivers a current of 5.0 A for 10 seconds, how many electrons flow through the device?
I = Q/t
Q = n*e
I = n*e/t --> n = I*t/e
n = 3.125*1020 electrons
As more resistors are added in parallel across a constant voltage source, the power supplied by the source
a. decreases.
b. does not change.
c. increases.
c. increases
P = V2/R
If you add resistors in parallel, V will remain constant. Power will only depend on resistance.
If you add more resistors in parallel, resistance will decrease, meaning power will increase because the two are inversely proportional.
A charge of 2.00 μC flows onto the plates of a capacitor when it is connected to a 12.0V potential source. What is the minimum amount of work that must be done in charging this capacitor?
C = Q/V
C = 2*10-6/12 = 1.66μF
U = 0.5*C*V2
U = 0.5*1.66*10-6*(122)
U = 12μJ
A parallel-plate capacitor consists of two parallel, square plates that have dimensions 1.0 cm by 1.0 cm. If the plates are separated by 1.0 mm, and the space between them is filled with teflon, what is the capacitance of this capacitor? (The dielectric constant for teflon is 2.1, and ε0 = 8.85 × 10-12)
C = (ε0*𝛋*A)/d
C = ((8.85*10-12)*2.1*(1*10-4)/(1*10-3)
C = 1.86*10-12 F
In a certain electroplating process gold is deposited by using a current of 14.0 A for 19 minutes. A gold ion, Au+, has a mass of approximately 3.3 × 10-22 g. How many grams of gold are deposited by this process?
33 g
I = Q/t --> Q = I*t = 15960 C
q = n*e --> n = q/e
n = 15960 C / (1.6*10-19C) = 9.975*1022 electrons
MAu = (9.975*1022)*(3.3 × 10-22 g) = 33g
A cylindrical wire has a resistance R and resistivity ρ. If its length and diameter are BOTH cut in half, what will be its resistance (in terms of R)? What will be its resistivity (in terms of ρ)?
2R and ρ
R = ρ*l/A (if both 1 and d are cut in half, the new resistance will be twice as big)
ρ = me/n*q2*𝛕 (does not depend on l or d)
An isolated air-filled parallel-plate capacitor that is no longer connected to anything has been charged up to Q = 2.9 nC. The separation between the plates initially is 1.20 mm, and for this separation the capacitance is 31 pF. Calculate the work that must be done to pull the plates apart until their separation becomes 5.30 mm, if the charge on the plates remains constant.
C = (ε0*A)/d --> A = C*d/ε0
A = 4.2*10-3
U1 = 0.5*Q2/C = 1.36*10-7J
C2 = (ε0*A)/d = (8.85*10-12)*(4.2*10-3)/(5.3*10-3) = 7*10-12 F
U2 = 0.5*Q2/C = 0.5*(1.362)/(7*10-12)
U2 = 6*10-7J
W = U2-U1 = 4.6*10-7J
A 15-μF air-filled capacitor is connected to a 50-V voltage source and becomes fully charged. The voltage source is then removed and a slab of dielectric that completely fills the space between the plates is inserted. The dielectric has a dielectric constant of 5.0.
(a) What is the capacitance of the capacitor after the slab has been inserted?
(b) What is the potential difference across the plates of the capacitor after the slab has been inserted?
E = Q /ε0*A --> Q = E*ε0*A
C = Q/V = E*ε0*A/E*d = ε0*A/d
(a) C1 = 𝛋*C0 = 5*(15*10-6) = 75μF
Q1 = 𝛋*Q0 (proportional to C)
V1 = V0/𝛋
(b) V1 = 50V/5 = 10V
When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now replaced with a silver wire having twice the diameter of the original wire, how much current will flow through the silver wire? The lengths of both wires are the same, and the voltage difference remains unchanged. (The resistivity of the original metal is 1.68*10-8 ohm*m, and the resistivity of silver is 1.59*10-8 ohm*m.)
a) 5.3 mA
b) 11 mA
c) 19 mA
d) 21 mA
d) 21 mA
R = ρ*l/A
Rs = Rm*(dm/ds)2*ρs/ρm (I cancels out)
Rs = Rm*(1/4)*ρs/ρm (I cancels out)
Rm/Rs = 4.23
I = Im*Rm/Rs
I = (5*10-3)*(4.23)
I = 21mA
A wire has a cross-sectional area of 0.10 mm2. If there are 4.0 × 1028 atoms per cubic meter in this wire, and if each atom contributes 2 free electrons, what is the drift velocity of the electrons when the current in the wire is 6.0 A?
A = 0.1mm2*(1m/1000mm)2 = 1*10-7m2
I = vd*n*q*A
vd = I/n*q*A = 6/(4*1028*2*1.6*10-19*10-7)
vd = 4.69*10-3 m/s