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Electrostatic Potential and Capacitance Test - 20

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Electrostatic Potential and Capacitance Test - 20
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Weekly Quiz Competition
  • Question 1
    1 / -0

    A proton is accelerated through 70,000 V. Its energy will increase by

    Solution

    Kinetic energy K = Q.V ⇒ K = (+e) (70000V)

    = 70000 eV = \(70000 \times 1.6 \times 10^{19}J\)

    \(11200 \times 10^{15}J\)

  • Question 2
    1 / -0

    The work done in bringing a 40 coulomb charge from point A to point B for distance 0.4 m is 5J. The potential difference between the two points will be (in volt)

    Solution

    By using W = \(Q. \triangle V\)

    \(\triangle V = \frac{5}{40}\)

    = 0.125 volt

  • Question 3
    1 / -0

    A parallel plate condenser has a capacitance of 60 µF in air and 200 µF when immersed in oil. The dielectric constant ′k′ of the oil is

    Solution

    \(C_{medium} = KC_{air}\)

    \(K = \frac{C_{medium}}{C_{air}}\)

    \(\frac{200}{60}\)

    = 3.33

  • Question 4
    1 / -0

    Assertion: If the distance between parallel plates of a capacitor is halved and dielectric constant is made three times, then the capacitor becomes 6 times.

    Reason: Capacity of the capacitor does not depend upon the nature of the material.

    (A) If both assertion and reason are true and the reason is the correct explanation of the assertion.

    (B) If both assertion and reason are true but reason is not the correct explanation of the assertion.

    (C) If assertion is true but reason is false.

    (D) If the assertion and reason both are false.

    (E) If assertion is false but reason is true

    Solution

    By the formula capacitance of a capacitor

    \(C_1 = \varepsilon_0 \times \frac{KA}{d} \propto \frac{K}{d}\)

    \(\frac{C_1}{C_2} = \frac{K_1}{d_1} \times \frac{d_2}{K_2}\)

    \(\frac{K_1}{K_2} \times \frac{\frac{d}{2}}{3K} = \frac{1}{6}\)

    or \(C_2 = 6 C_1\)

    C = \(\frac{Q}{V}\)

    Again for capacity of a capacitor C = \(\frac{Q}{V}\). Therefore, capacity of a capacitor does not depend upon the nature of the material of the capacitor.

  • Question 5
    1 / -0

    Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is

    Solution

    Volume of 8 small drops = Volume of big drop

    \(8 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3\)

    R = 2r,  hence capacity becomes 2 times.

  • Question 6
    1 / -0

    The distance between the circular plates of a parallel plate condenser 40 mm in diameter, in order to have same capacity as a sphere of radius 1 metres

    Solution

    \(4 \pi \varepsilon_0 r = \frac{\varepsilon_0 A}{d}\)

    d = \(\frac{A}{4 \pi r} = \frac{\pi (20 \times 10^{-3})^2}{4 \pi \times 1}\)

    = 0.1 mm

  • Question 7
    1 / -0

    The capacitance of a metallic sphere will be 1µF, if its radius is nearly

    Solution

    \(4 \pi \varepsilon_0 r = 1 \times 10^{-6}\)

    r = \(10^{-6} \times 9 \times 10^9\)

    = 9 km

  • Question 8
    1 / -0

    A parallel plate capacitor of a capacitance of 1 farad would have the plate area about

    Solution

    Capacitance, C = 1 F

    Let 

    d = 1 pm

    \(1 \times 10^{-12} m\)

    Capacitance,

    \(C = \frac{\varepsilon_0 A}{d} \)

    \(A = \frac{Cd}{\varepsilon_0}\)

    \(\frac{1 \times 1 \times 10^{-12}}{8.854 \times 10^{-12}}\)

    \(= 0.113 m^2\)

  • Question 9
    1 / -0

    Energy stored in capacitor and dissipated during charging a capacitor bear a ratio

    Solution

    Half of the energy is dissipated during charging a capacitor.

    Hence, \(\frac{energy\;stored\;in\;a\;capacitor}{energy\;dissipated\;during\;charging\;a\;capacitor}\)

    \(\frac{2}{1}\)

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