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Electrostatic Potential and Capacitance Test - 26

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Electrostatic Potential and Capacitance Test - 26
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  • Question 1
    1 / -0
    One Volt is equal to:
    Solution
    The volt is a measure of electric potential. One volt is defined as the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points. It is also equal to the potential difference between two parallel, infinite planes spaced 1 meter apart that create an electric field of 1 newton per coulomb. Additionally, it is the potential difference between two points that will impart one joule of energy per coulomb of charge that passes through it. It can be expressed in terms of SI base units (m, kg, s, and A)
    $$1 volt = \dfrac{1 Joule}{Coulomb}$$
  • Question 2
    1 / -0
    What is the S.I. unit of electric potential? 
    Solution
    An electric potential (also called the electric field potential or the electrostatic potential) is the amount of electric potential energy that a unitary point electric charge would have if located at any point of space, and is equal to the work done by an electric field in carrying a unit positive charge from infinity to that point.
    This value can be calculated in either a static (time-invariant) or a dynamic (varying with time) electric field at a specific time in units of joules per coulomb, or volts (V). The electric potential at infinity is assumed to be zero.
  • Question 3
    1 / -0
    Identify the correct statement from below:

    Solution
    Point A have positive charge.
    More is the positive charge, more will be the potential. So, A will be at higher potential.
  • Question 4
    1 / -0
    The electric volt is a measure of electrical potential. Identify which of the following can be defined as a volt.
    Solution
    eV is defined as the kinetic energy gained by a unit charge $$e$$ when it is accelerated by a potential difference of 1 volt. As we know kinetic energy is the measure of work done, therefore, volt can be defined as work done per unit charge.
  • Question 5
    1 / -0
    Two charges $$+q$$ and $$-q$$ are kept apart. Then at any point on the right bisector of line joining the two charges
    Solution
    At equatorial point
    $${ E }_{ e }=\dfrac { 1 }{ 4\pi { \varepsilon  }_{ 0 } } \dfrac { p }{ { r }^{ 3 } } $$
    (directed from $$+q$$ to $$-q$$) and $${ V }_{ e }=0$$
  • Question 6
    1 / -0
    Four capacitors of equal capacitance have an equivalent capacitance $${ C }_{ 1 }$$ when connected in series and an equivalent capacitance $${ C }_{ 2 }$$ when connected in parallel. The ratio $$\dfrac { { C }_{ 1 } }{ { C }_{ 2 } } $$ is
    Solution
    Let capacitance of each capacitor is $$C$$.
    In series, equivalent capacitance
        $${ C }_{ 1 }=\dfrac { C }{ 4 } $$
    In parallel, equivalent capacitance
        $${ C }_{ 2 }=4C$$
    $$\therefore \dfrac { { C }_{ 1 } }{ { C }_{ 2 } } =\dfrac { 1 }{ 16 } $$
  • Question 7
    1 / -0
    The intensity of electric field E due to charge Q at distance r.
    Solution
    According to Coulomb's Law: 
    $$E = k \frac{Q}{r^2}$$
    where E is electric field, q is charge at a distance r and k is a constant. Therefore, 
    $$E \propto \frac{1}{r^2}$$
  • Question 8
    1 / -0
    Identify the dimension of electric potential,
    Solution
    Work done $$=$$ charge $$\times $$ potential difference
    or Force $$\times$$ distance $$=$$ (current\times time)$$\times $$ potential difference
    or $$[MLT^{-2}]\times [L]=([IT])[V]$$
    or $$[V]=[ML^2T^{-3}I^{-1}]$$
  • Question 9
    1 / -0
    Two charges are separated by a distance d. If the distance between them is doubled, how does the electric potential between them change?
    Solution
    Initial electric potential between the charges         $$V=  \dfrac{kq}{d}$$
    New distance between the charges       $$d' = 2d$$

    Thus new electric potential         $$V' = \dfrac{kq}{2d}  =\dfrac{V}{2}$$
  • Question 10
    1 / -0
    What is the equivalent capacitance between $$A$$ and $$B$$ in the diagram below?

    Solution
    The equivalent capacitance of the two capacitors connected in parallel=$$3\mu F+2\mu F=5\mu F$$
    This combination is in series with $$5\mu F$$ capacitor.
    Hence final equivalent capacitance=$$\dfrac{5\times 5}{5+5}F=2.5F$$
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