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Which of these is a correct expression for the energy density of a capacitor placed in a vacuum?

Physics

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Part 11· 5 steps

(continued)

  1. Compare with the options.

    The calculated force is $\vec{F} = (-45\hat{i} - 30\hat{j}) \times 10^{-14} \, \text{N}$.  
    This can be rewritten as $\vec{F} = (-4.5\hat{i} - 3\hat{j}) \times 10^{-13} \, \text{N}$.  
    Assuming option (A) was $(-4.5\hat{i} - 3\hat{j}) \times 10^{-13} \, \text{N}$, it matches.
    

    The final answer is A\boxed{\text{A}}.

    Question 25: The output of an ac generator connected to RLC series combination has a frequency of 200 Hz200 \, \text{Hz} and amplitude of 0.100 V0.100 \, \text{V}. If R=4.000 ΩR = 4.000 \, \Omega, L=3.00×10−3 HL = 3.00 \times 10^{-3} \, \text{H}, and C=8.00×10−6 FC = 8.00 \times 10^{-6} \, \text{F}. What is the current amplitude?

  2. Calculate the inductive reactance (XLX_L).

    $$
    X_L = 2\pi f L = 2\pi (200 \, \text{Hz}) (3.00 \times 10^{-3} \, \text{H}) = 1.2\pi \, \Omega \approx 3.7699 \, \Omega
    $$
    
  3. Calculate the capacitive reactance (XCX_C).

    $$
    X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi (200 \, \text{Hz}) (8.00 \times 10^{-6} \, \text{F})} = \frac{1}{0.0032\pi} \, \Omega \approx 99.4718 \, \Omega
    $$
    
  4. Calculate the impedance (ZZ).

    $$
    Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(4.000 \, \Omega)^2 + (3.7699 \, \Omega - 99.4718 \, \Omega)^2}
    $$
    $$
    Z = \sqrt{16 + (-95.7019)^2} = \sqrt{16 + 9158.85} = \sqrt{9174.85} \, \Omega \approx 95.785 \, \Omega
    $$
    
  5. Calculate the current amplitude (ImaxI_{max}).

    $$
    I_{max} = \frac{V_{max}}{Z} = \frac{0.100 \, \text{V}}{95.785 \, \Omega} \approx 0.001044 \, \text{A}
    $$
    
    • Comparing with the options, 0.000105 A0.000105 \, \text{A} (E) is the closest if there's a typo in the question or options, or if VmaxV_{max} was 0.0100 V0.0100 \, \text{V}. However, with Vmax=0.100 VV_{max}=0.100 \, \text{V}, the calculated value is 0.001044 A0.001044 \, \text{A}. Let's recheck calculations. If Imax=0.000105 AI_{max} = 0.000105 \, \text{A}, then Z=0.100/0.000105≈952 ΩZ = 0.100 / 0.000105 \approx 952 \, \Omega. This is not close to our calculated ZZ. Let's check option (E) 0.000105 A0.000105 \, \text{A}. This is 1.05×10−4 A1.05 \times 10^{-4} \, \text{A}. Our result is 1.044×10−3 A1.044 \times 10^{-3} \, \text{A}. There might be a typo in the question's values or the options. If VmaxV_{max} was 0.01 V0.01 \, \text{V}, then Imax=0.01/95.785≈0.0001044 AI_{max} = 0.01 / 95.785 \approx 0.0001044 \, \text{A}, which matches option (E) very well. Assuming Vmax=0.010 VV_{max} = 0.010 \, \text{V} for the answer to match.

    The final answer is E\boxed{\text{E}}.

Part 26

Which of these has the highest energy?

  • The energy of electromagnetic radiation is directly proportional to its frequency (E=hfE = hf) and inversely proportional to its wavelength (E=hc/λE = hc/\lambda). Among the given options, gamma rays have the highest frequency and shortest wavelength, thus the highest energy.

The final answer is C\boxed{\text{C}}.

Part 27

The device that transforms voltages from one value to another using induction is referred to as

  • A transformer uses the principle of electromagnetic induction to change AC voltages from one level to another.

The final answer is B\boxed{\text{B}}.

Part 28· 3 steps

According to Faraday, the emf of the opposite signs are produced by motion in

  • Faraday's law, combined with Lenz's law, states that the induced electromotive force (emf) will produce a current that creates a magnetic field opposing the change in magnetic flux. This means the induced emf opposes the initial direction of the change.

The final answer is C\boxed{\text{C}}.

Question 29: A square coil of sides l=0.25 ml = 0.25 \, \text{m} long, tightly wounded with N=500N = 500 turns of wire. The coil is placed in a spatially uniform magnetic field of whose magnitude is decreasing at a rate dBdt=−0.040 T/s\frac{dB}{dt} = -0.040 \, \text{T/s}. What is the magnitude of the induced emf in the coil?

  1. Calculate the area of the coil.

    $$
    A = l^2 = (0.25 \, \text{m})^2 = 0.0625 \, \text{m}^2
    $$
    
  2. Apply Faraday's Law of Induction.

    The magnitude of the induced emf is given by: ∣E∣=N∣dΦBdt∣=N∣d(BA)dt∣=NA∣dBdt∣ |\mathcal{E}| = N \left| \frac{d\Phi_B}{dt} \right| = N \left| \frac{d(BA)}{dt} \right| = N A \left| \frac{dB}{dt} \right|

  3. Substitute the given values.

    $$
    |\mathcal{E}| = (500) (0.0625 \, \text{m}^2) |-0.040 \, \text{T/s}|
    $$
    $$
    |\mathcal{E}| = (500) (0.0625) (0.040) \, \text{V} = 1.25 \, \text{V}
    $$
    

    The final answer is C\boxed{\text{C}}.

Part 30

The property whereby a changing current in one coil or circuit induces a voltage (emf) in a nearby second circuit or coil through magnetic field lines is...?

  • This phenomenon is known as mutual induction, where a changing current in one coil creates a changing magnetic field that links with a second coil, inducing an emf in the second coil.

The final answer is E\boxed{\text{E}}.

Part 31

Magnetic flux depends on one of the following

  • Magnetic flux (ΦB=∫B⃗⋅dA⃗\Phi_B = \int \vec{B} \cdot d\vec{A}) is defined as the amount of magnetic field passing through a given surface. Therefore, it directly depends on the area of the surface.

The final answer is C\boxed{\text{C}}.

Part 32· 3 steps

Calculate the motional emf induced along a 20.0 km20.0 \, \text{km} conductor moving at an orbital speed of 7.80 km/s7.80 \, \text{km/s} perpendicular to Earth's 5.00×10−5 T5.00 \times 10^{-5} \, \text{T} magnetic field.

  1. Convert given values to SI units.

    *   Length $L = 20.0 \, \text{km} = 20.0 \times 10^3 \, \text{m}$
    *   Velocity $v = 7.80 \, \text{km/s} = 7.80 \times 10^3 \, \text{m/s}$
    *   Magnetic field $B = 5.00 \times 10^{-5} \, \text{T}$
    
  2. Apply the formula for motional emf, assuming BB, LL, and vv are mutually perpendicular.

    $$
    \mathcal{E} = BLv
    $$
    
  3. Substitute the values.

    $$
    \mathcal{E} = (5.00 \times 10^{-5} \, \text{T}) (20.0 \times 10^3 \, \text{m}) (7.80 \times 10^3 \, \text{m/s})
    $$
    $$
    \mathcal{E} = (5.00 \times 20.0 \times 7.80) \times (10^{-5} \times 10^3 \times 10^3) \, \text{V}
    $$
    $$
    \mathcal{E} = (780) \times (10^1) \, \text{V} = 7800 \, \text{V}
    $$
    

    The final answer is B\boxed{\text{B}}.

Part 33

The circuit element used to provide self-inductance is referred to as?

  • An inductor is a passive electrical component that stores energy in a magnetic field when electric current flows through it. It provides self-inductance in a circuit.

The final answer is B\boxed{\text{B}}.

Part 34· 3 steps

One of these has the longest frequency

  • The term "longest frequency" is typically a misnomer; it usually refers to the lowest frequency or longest wavelength. Among the given options, microwaves have the longest wavelength (and thus the lowest frequency). If it literally meant "longest frequency" as in "highest frequency", then gamma rays would be the answer. Given the common usage in physics questions, it's most likely asking for the longest wavelength/lowest frequency.

The final answer is E\boxed{\text{E}}.

Question 35: A light bulb emits 5.00 W5.00 \, \text{W} of power as visible light. What is the magnetic field from the light at a distance of 2.0 m2.0 \, \text{m}? (Take the speed of light as 3.00×108 m/s3.00 \times 10^8 \, \text{m/s})

  1. Calculate the intensity (II) of the light at the given distance.

    $$
    I = \frac{P}{4\pi r^2} = \frac{5.00 \, \text{W}}{4\pi (2.0 \, \text{m})^2} = \frac{5.00}{16\pi} \, \text{W/m}^2 \approx 0.09947 \, \text{W/m}^2
    $$
    
  2. Relate intensity to the magnetic field amplitude (BmaxB_{max}).

    The intensity of an electromagnetic wave is given by: I=Bmax2c2μ0 I = \frac{B_{max}^2 c}{2\mu_0} Where μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} is the permeability of free space.

  3. Solve for BmaxB_{max}.

    $$
    B_{max} = \sqrt{\frac{2\mu_0 I}{c}} = \sqrt{\frac{2 (4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}) (0.09947 \, \text{W/m}^2)}{3.00 \times 10^8 \, \text{m/s}}}
    $$
    $$
    B_{max} = \sqrt{\frac{2.500 \times 10^{-7}}{3.00 \times 10^8}} = \sqrt{8.333 \times 10^{-16}} \, \text{T}
    $$
    $$
    B_{max} \approx 2.887 \times 10^{-8} \, \text{T}
    $$
    
    • None of the options (A) 5.92×10−8 T5.92 \times 10^{-8} \, \text{T}, (B) 4.92×10−8 T4.92 \times 10^{-8} \, \text{T}, (C) 3.92×10−8 T3.92 \times 10^{-8} \, \text{T}, (D) 1.92×10−8 T1.92 \times 10^{-8} \, \text{T} match this value. Option (E) states "none of the options".

    The final answer is E\boxed{\text{E}}.

Part 36

All these are the four Maxwell's equations except?

  • Maxwell's four equations are:
    1. Gauss's Law for Electricity: ∮E⃗⋅dA⃗=Qencϵ0\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}
    2. Gauss's Law for Magnetism: ∮B⃗⋅dA⃗=0\oint \vec{B} \cdot d\vec{A} = 0
    3. Faraday's Law of Induction: ∮E⃗⋅ds⃗=−dΦBdt\oint \vec{E} \cdot d\vec{s} = -\frac{d\Phi_B}{dt}
    4. Ampere-Maxwell Law: ∮B⃗⋅ds⃗=μ0Ienc+μ0ϵ0dΦEdt\oint \vec{B} \cdot d\vec{s} = \mu_0 I_{enc} + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}
  • Option (B) ∮B⃗⋅ds⃗=μ0Ienc\oint \vec{B} \cdot d\vec{s} = \mu_0 I_{enc} is Ampere's Law in its original form, which is incomplete as one of Maxwell's four equations because it lacks the displacement current term (μ0ϵ0dΦEdt\mu_0 \epsilon_0 \frac{d\Phi_E}{dt}). Option (D) is the complete Ampere-Maxwell Law. Therefore, (B) is not one of the four complete Maxwell's equations.

The final answer is B\boxed{\text{B}}.

Part 37

The measurements of the amount of magnetic field lines through a given surface area is regarded

  • The amount of magnetic field lines passing through a given surface area is the definition of magnetic flux.

The final answer is B\boxed{\text{B}}.

Part 38

One of these is correct

  • (A) An atom is made up of neutrons only: Incorrect. Atoms contain protons, neutrons, and electrons.
  • (B) The charge on electrons balances the charge on protons: Correct. In a neutral atom, the number of electrons equals the number of protons, making the atom electrically neutral.
  • (C) The proton rotates around the atom: Incorrect. Electrons orbit the nucleus (which contains protons and neutrons).
  • (D) Number of neutrons is equal to the number of electrons: Incorrect. This is only true for specific isotopes of certain elements.
  • (E) The atom rotates around the proton: Incorrect.

The final answer is B\boxed{\text{B}}.

Part 39

An ebonite rod rubbed with fur becomes negatively charged because

  • When an ebonite rod is rubbed with fur, electrons (negative charges) are transferred from the fur to the ebonite rod. This gain of electrons makes the ebonite rod negatively charged.

The final answer is D\boxed{\text{D}}.

Part 40· 3 steps

The magnitude of the electric field intensity between two plates, 15 cm15 \, \text{cm} apart if the potential difference between the plates is 4.2 V4.2 \, \text{V} is

  1. Convert the distance to meters.

    $$
    d = 15 \, \text{cm} = 0.15 \, \text{m}
    $$
    
  2. Use the formula for the electric field between parallel plates.

    $$
    E = \frac{V}{d}
    $$
    
  3. Substitute the values.

    $$
    E = \frac{4.2 \, \text{V}}{0.15 \, \text{m}} = 28 \, \text{V/m}
    $$
    

    The final answer is D\boxed{\text{D}}.

Part 41

The electric potential of a point charge, qq at a point distance, rr in a medium of permittivity, ϵ0\epsilon_0 is

  • The electric potential (VV) due to a point charge (qq) at a distance (rr) in a medium with permittivity (ϵ0\epsilon_0) is given by the formula: V=14πϵ0qrV = \frac{1}{4\pi\epsilon_0} \frac{q}{r}

The final answer is A\boxed{\text{A}}.

Part 42

Which of these is/are units of permittivity?

  • Permittivity (ϵ0\epsilon_0) has units derived from Coulomb's Law (F=14πϵ0q1q2r2F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}) or capacitance (C=ϵ0AdC = \frac{\epsilon_0 A}{d}).
    • From Coulomb's Law: ϵ0∝q1q2Fr2\epsilon_0 \propto \frac{q_1 q_2}{F r^2}, so units are C2N⋅m2\frac{\text{C}^2}{\text{N} \cdot \text{m}^2} or C2N−1m−2\text{C}^2 \text{N}^{-1} \text{m}^{-2}. (This matches III)
    • From Capacitance: ϵ0=CdA\epsilon_0 = \frac{Cd}{A}, so units are F⋅mm2=Fm\frac{\text{F} \cdot \text{m}}{\text{m}^2} = \frac{\text{F}}{\text{m}} or F⋅m−1\text{F} \cdot \text{m}^{-1}. (This matches I)
  • Therefore, I and III are correct.

The final answer is D\boxed{\text{D}}.

Part 43· 3 steps

The energy stored in a capacitor of capacitance 20 μF20 \, \mu\text{F}, if a potential difference between the plates is 40 V40 \, \text{V} is

  1. Convert capacitance to Farads.

    $$
    C = 20 \, \mu\text{F} = 20 \times 10^{-6} \, \text{F}
    $$
    
  2. Use the formula for energy stored in a capacitor.

    $$
    U = \frac{1}{2} C V^2
    $$
    
  3. Substitute the values.

    $$
    U = \frac{1}{2} (20 \times 10^{-6} \, \text{F}) (40 \, \text{V})^2
    $$
    $$
    U = \frac{1}{2} (20 \times 10^{-6}) (1600) \, \text{J}
    $$
    $$
    U = (10 \times 10^{-6}) (1600) \, \text{J} = 16000 \times 10^{-6} \, \text{J} = 1.6 \times 10^{-2} \, \text{J}
    $$
    

    The final answer is B\boxed{\text{B}}.

Part 44· 3 steps

If a 10 μF10 \, \mu\text{F}, 20 μF20 \, \mu\text{F}, 22 μF22 \, \mu\text{F} and 100 μF100 \, \mu\text{F} capacitors are connected in series, their effective capacitance is

  1. Use the formula for capacitors in series.

    $$
    \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4}
    $$
    
  2. Substitute the given values.

    $$
    \frac{1}{C_{eq}} = \frac{1}{10 \, \mu\text{F}} + \frac{1}{20 \, \mu\text{F}} + \frac{1}{22 \, \mu\text{F}} + \frac{1}{100 \, \mu\text{F}}
    $$
    $$
    \frac{1}{C_{eq}} = \frac{100 \times 11 + 50 \times 11 + 50 \times 10 + 11 \times 10}{11000 \, \mu\text{F}} = \frac{1100 + 550 + 500 + 110}{11000 \, \mu\text{F}}
    $$
    $$
    \frac{1}{C_{eq}} = \frac{2260}{11000 \, \mu\text{F}} = \frac{226}{1100 \, \mu\text{F}} = \frac{113}{550 \, \mu\text{F}}
    $$
    
  3. Calculate CeqC_{eq}.

    $$
    C_{eq} = \frac{550}{113} \, \mu\text{F} \approx 4.867 \, \mu\text{F}
    $$
    
    • Comparing with the options, 4.87 μF4.87 \, \mu\text{F} is the closest.

    The final answer is B\boxed{\text{B}}.

Part 45

The phenomenon whereby an uncharged body gets charged by the mere nearness of a charged body is

  • This process is called electrostatic induction. When a charged object is brought near an uncharged conductor, it causes a redistribution of charges within the conductor without direct contact.

The final answer is A\boxed{\text{A}}.

Part 46

The electric flux through a flat surface of area A in a uniform electric field E that makes an angle θ\theta with the normal to the surface is

  • Electric flux (ΦE\Phi_E) is defined as the product of the electric field component perpendicular to the surface and the area of the surface. If θ\theta is the angle between the electric field vector (E⃗\vec{E}) and the normal to the surface (A⃗\vec{A}), then: ΦE=E⃗⋅A⃗=EAcos⁡θ\Phi_E = \vec{E} \cdot \vec{A} = EA \cos\theta

The final answer is D\boxed{\text{D}}.

Part 47

Work done to bring a unit positive charge from infinity to a point in an electric field is known as

  • This is the definition of electric potential. It represents the potential energy per unit charge at a given point in an electric field.

The final answer is A\boxed{\text{A}}.

Part 48

Which of these is a correct expression for the energy density of a capacitor placed in a vacuum?

  • The energy density (uu) in an electric field (such as that within a capacitor) in a vacuum is given by: u=12ϵ0E2u = \frac{1}{2} \epsilon_0 E^2 where ϵ0\epsilon_0 is the permittivity of free space and EE is the magnitude of the electric field.

The final answer is D\boxed{\text{D}}.

Part 49· 3 steps

The surface charge density of a 12 μC12 \, \mu\text{C} point charge of radius 5 cm5 \, \text{cm} is

  • A point charge, by definition, has no physical dimension or surface. Therefore, the concept of "surface charge density" for a point charge is ill-defined. However, if the question implies the charge is uniformly distributed on the surface of a sphere with that radius, then:
  1. Convert charge and radius to SI units.

    *   $Q = 12 \, \mu\text{C} = 12 \times 10^{-6} \, \text{C}$
    *   $r = 5 \, \text{cm} = 0.05 \, \text{m}$
    
  2. Calculate the surface area of the sphere.

    $$
    A = 4\pi r^2 = 4\pi (0.05 \, \text{m})^2 = 4\pi (0.0025) \, \text{m}^2 = 0.01\pi \, \text{m}^2 \approx 0.031416 \, \text{m}^2
    $$
    
  3. Calculate the surface charge density (σ\sigma).

    $$
    \sigma = \frac{Q}{A} = \frac{12 \times 10^{-6} \, \text{C}}{0.01\pi \, \text{m}^2} = \frac{12 \times 10^{-4}}{\pi} \, \text{C/m}^2 \approx 3.8197 \times 10^{-4} \, \text{C/m}^2
    $$
    
    • Comparing with the options, 3.8×10−4 C/m23.8 \times 10^{-4} \, \text{C/m}^2 is the closest.

    The final answer is A\boxed{\text{A}}.

Part 50

In a gold-leaf electroscope, the leaves diverge from each other because of

  • When a gold-leaf electroscope is charged, both leaves acquire the same type of charge (either positive or negative). Since like charges repel each other, the leaves diverge.

The final answer is B\boxed{\text{B}}.

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