Energy And Electricity Quiz

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  • 1/13 Questions

    A metal sphere of radius 2.0 cm carries a charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sphere?

    • 3.4 × 10^6 N/C
    • 9.3 × 10^6 N/C
    • 7.5 × 10^6 N/C
    • 4.2 × 10^6 N/C
    • 5.7 × 10^6 N/C
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About This Quiz

This quiz covers basic concepts related to electricity, such as capacitance, ohm's law, inductance, electric force, and electric potential.

Energy And Electricity Quiz - Quiz

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  • 2. 

    An infinitely long cylinder of radius R = 2 cm carries a uniform charge density Sigma = 18 μC/m3. Calculate the electric field at distance r = 1 cm from the axis of the cylinder.

    • 0

    • 2.5 × 103 N/C

    • 10.2 × 103 N/C

    • 5.1 × 103 N/C

    • 2.0 × 103 N/C

    Correct Answer
    A. 10.2 × 103 N/C
    Explanation
    The electric field at a distance r from the axis of an infinitely long cylinder with uniform charge density can be calculated using the formula E = (Sigma * r)/(2 * epsilon), where Sigma is the charge density, r is the distance from the axis, and epsilon is the permittivity of free space. Plugging in the given values, we get E = (18 μC/m3 * 1 cm)/(2 * epsilon). Simplifying this expression gives E = 9 μC/(2 * epsilon). Using the value of epsilon, which is approximately 8.85 x 10^-12 C2/(N*m2), we can calculate E to be approximately 10.2 x 103 N/C.

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  • 3. 

    The plates of a parallel-plate capacitor are maintained with constant voltage by a battery as they are pulled apart. What happens to the strength of the electric field during this process?

    • It remains constant.

    • It increases.

    • It decreases.

    • Cannot be determined from the information given.

    Correct Answer
    A. It decreases.
    Explanation
    As the plates of a parallel-plate capacitor are pulled apart, the distance between them increases. The strength of the electric field between the plates is directly proportional to the voltage applied and inversely proportional to the distance between the plates. Therefore, as the distance increases, the strength of the electric field decreases.

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  • 4. 

    A positive object touches a neutral electroscope, and the leaves separate. Then a negative object is brought near the electroscope, but does not touch it. What happens to the leaves?

    • They are unaffected.

    • They separate further.

    • They move closer together.

    • Cannot be determined without further information.

    Correct Answer
    A. They move closer together.
    Explanation
    When a positive object touches a neutral electroscope, the leaves separate because the positive charges are repelled and move away from each other. However, when a negative object is brought near the electroscope, the negative charges attract the positive charges in the electroscope, causing them to move closer together. Therefore, the correct answer is "They move closer together."

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  • 5. 

    If the electric field is 12 V/m in the positive x-direction, what is the potential difference between the origin, (0,0), and the point (3.0 m, 4.0 m)?

    • 48 V with the origin at the lower potential

    • 36 V with the origin at the higher potential

    • 48 V with the origin at the higher potential

    • 36 V with the origin at the lower potential

    • Cannot be determined

    Correct Answer
    A. 36 V with the origin at the higher potential
    Explanation
    The potential difference between two points in an electric field is given by the formula V = Ed, where V is the potential difference, E is the electric field, and d is the distance between the two points. In this case, the electric field is given as 12 V/m in the positive x-direction. The distance between the origin (0,0) and the point (3.0 m, 4.0 m) can be calculated using the Pythagorean theorem, which gives a distance of 5.0 m. Plugging these values into the formula, we get V = (12 V/m)(5.0 m) = 60 V. However, since the question asks for the potential difference with the origin at the higher potential, the answer is 36 V.

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  • 6. 

    Two long straight parallel lines of charge, #1 and #2, carry positive charge per unit lengths of Lambda1 and Lambda2, respectively. Lambda1 > Lambda2. The locus of points where the electric field is zero in this case is:

    • Along a line between the lines closer to line #2 than line #1.

    • Along a line between the lines closer to line #1 than line #2.

    • At a point halfway between the lines.

    • Along line #1.

    • Cannot be determined.

    Correct Answer
    A. Along a line between the lines closer to line #2 than line #1.
    Explanation
    The electric field due to a line of charge is inversely proportional to the distance from the line. Since Lambda1 is greater than Lambda2, the electric field due to line #1 is stronger than the electric field due to line #2. Therefore, there must be a point between the lines where the electric fields due to line #1 and line #2 cancel each other out, resulting in zero net electric field. This point will be closer to line #2 than line #1. Hence, the locus of points where the electric field is zero is along a line between the lines closer to line #2 than line #1.

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  • 7. 

    A negative charge, if free, tries to move:

    • Away from infinity.

    • From high potential to low potential.

    • In the direction of the electric field.

    • From low potential to high potential.

    • Toward infinity.

    Correct Answer
    A. From low potential to high potential.
    Explanation
    A negative charge, being attracted to positive charges, naturally moves from areas of low potential (where there are fewer positive charges) to areas of high potential (where there are more positive charges). This movement allows the negative charge to reduce the overall potential energy of the system and reach a more stable state. Therefore, a negative charge tries to move from low potential to high potential.

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  • 8. 

    Which of the following cylindrical wires has the largest resistance? All wires are made of the same material.

    • A wire of length L and diameter d

    • A wire of length L and diameter d/2

    • A wire of length L/2 and diameter 2d

    • A wire of length L/2 and diameter d

    • A wire of length L and diameter 2d

    Correct Answer
    A. A wire of length L and diameter d/2
    Explanation
    The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. In this case, all wires are made of the same material, so the resistivity is constant.

    Comparing the wires, we can see that the wire with the largest resistance will have the longest length and the smallest diameter.

    Therefore, the wire of length L and diameter d/2 will have the largest resistance because it has the longest length and the smallest diameter among the given options.

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  • 9. 

    A dielectric is inserted between the plates of an isolated parallel-plate capacitor that carries a charge Q. What happens to the potential energy stored in the capacitor?

    • The potential energy of the capacitor decreases.

    • The potential energy of the capacitor remains the same.

    • The potential energy of the capacitor increases.

    • The potential energy of the capacitor increases or decreases depending on the value of the dielectric constant of the capacitor.

    • More information is needed to answer the question.

    Correct Answer
    A. The potential energy of the capacitor decreases.
    Explanation
    When a dielectric is inserted between the plates of a parallel-plate capacitor, it increases the capacitance of the capacitor. As the potential energy stored in a capacitor is directly proportional to the capacitance, the potential energy decreases when the capacitance increases. Therefore, the potential energy of the capacitor decreases when a dielectric is inserted between the plates.

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  • 10. 

    Two charges QA = +q and QB = - 3q are located on the x-axis at x= 0 and x= d respectively. Where is the electric potential equal to zero?

    • X = d/4

    • X= d/2

    • X= d/3

    • X= 2d/3

    • X= 3d/4

    Correct Answer
    A. X = d/4
    Explanation
    The electric potential is zero at a point on the x-axis when the sum of the electric potential due to the positive charge QA and the electric potential due to the negative charge QB is equal to zero. In this case, the electric potential due to QA is positive and the electric potential due to QB is negative. Since QB has a greater magnitude, the electric potential due to QB will be larger. Therefore, the point where the electric potential is zero will be closer to QB. Since QB is located at x = d and the point x = d/4 is closer to QB than any other option, the electric potential is equal to zero at x = d/4.

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  • 11. 

    A 500-W device is connected to a 120-V ac power source. What peak current flows through this device?

    • 170 A

    • 5.9 A

    • 4.2 A

    • 150 A

    • 120 A

    Correct Answer
    A. 5.9 A
    Explanation
    The peak current flowing through a device can be calculated using the formula I = P/V, where I is the current, P is the power, and V is the voltage. In this case, the power is 500 W and the voltage is 120 V. Plugging these values into the formula, we get I = 500/120 = 4.2 A. Therefore, the correct answer is 4.2 A.

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  • 12. 

    A uniform electric field E = Esubzero j is set-up along the y-axis in a region of space. A frame is placed in that region in such a way that its plane is perpendicular to the y-axis. Which of the following changes would decrease the magnitude of the electric flux through the frame?

    • Sliding the frame sideways parallel to the z-axis within the xz-plane.

    • Moving the frame vertically along the y-axis keeping parallel to the xz-plane.

    • Sliding the frame sideways parallel to the x-axis within the xz-plane.

    • Tilting the frame so that its plane is now in the yz-plane.

    • Rotating the frame in the xz-plane with respect to the y-axis.

    Correct Answer
    A. Tilting the frame so that its plane is now in the yz-plane.
    Explanation
    Tilting the frame so that its plane is now in the yz-plane would decrease the magnitude of the electric flux through the frame. This is because the electric field is set up along the y-axis, and when the frame is tilted to the yz-plane, it will no longer be perpendicular to the electric field. The electric flux is maximized when the frame is perpendicular to the electric field, so any change that deviates from this orientation will decrease the magnitude of the electric flux.

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  • 13. 

    A 6.00-μF parallel plate capacitor has a charge of +40.0 μC and -40.0 μC on each plate, respectively. The potential energy stored in this capacitor is:

    • 123 μJ

    • 113 μJ

    • 103 μJ

    • 143 μJ

    • 133 μJ

    Correct Answer
    A. 133 μJ
    Explanation
    The potential energy stored in a capacitor is given by the formula U = (1/2) * C * V^2, where U is the potential energy, C is the capacitance, and V is the voltage across the capacitor. In this case, the capacitance is given as 6.00 μF. The voltage can be calculated using the formula V = Q / C, where Q is the charge on the capacitor. Since the charge on each plate is +40.0 μC and -40.0 μC, the total charge is 80.0 μC. Plugging these values into the formula, we get V = 80.0 μC / 6.00 μF = 13.3 V. Finally, substituting the values of C and V into the formula for potential energy, we get U = (1/2) * 6.00 μF * (13.3 V)^2 = 133 μJ.

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  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Oct 13, 2010
    Quiz Created by
    Jwm9q
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