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Electrostatic Potential and Capacitance Test - 27

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Electrostatic Potential and Capacitance Test - 27
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  • Question 1
    1 / -0
    When moving electron comes closer to other stationary electron, then its kinetic energy and potential energy respectively _____ and _____.
    Solution
    When electron comes closer to the other stationary electron, its kinetic energy decreases because of repulsion between them. As per conservation of energy, the potential energy increases.
  • Question 2
    1 / -0
    The equivalent capacitance between the points A and B in the given diagram is:

    Solution
    $${C_S}$$ Capacitance in series
    $$\begin{array}{l} \dfrac { 1 }{ { { C_{ S } } } } =\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 }  \\ =\dfrac { 3 }{ 2 }  \\ { C_{ S } }=\dfrac { 2 }{ 3 }  \end{array}$$
    The equivalent capacitance between A & B is 
    $$\begin{array}{l} { C_{ P } }=\dfrac { 2 }{ 3 } +2 \\ { C_{ P } }=\dfrac { 8 }{ 3 } \mu F \end{array}$$

  • Question 3
    1 / -0
    From a supply of identical capacitors rated $$8\ \mu F, 250\ V$$, the minimum number of capacitors required to form a composite $$16\ \mu F, 1000\ V$$ capacitor is:
    Solution
    The required voltage is 1000v and the capacitors are parallel as 250v.
    So number of capacitors required will be 4  i.e$$ 250 \times 4 = 1000$$ in series.
    Now example of four capacitor in series will be equal $$2 \mu f$$ (micro farade) but the equivalent capacitance required is given as $$16\mu f$$ so there must be 8 series of parallel arrange capacitors each of capacitor 2 micro farad hence total number of capacitor $$= 4 \times 8 = 32$$

    Hence option (D) is corrct
  • Question 4
    1 / -0
    The dielectric constant of air is.
    Solution
    Dielectric constant of a medium $$K = \dfrac{\epsilon}{\epsilon_o}$$
    where $$\epsilon$$ is the permittivity of the medium.
    For air, $$\epsilon = \epsilon_o$$
    $$\implies$$ $$K = 1$$
  • Question 5
    1 / -0
    What is not true of equipotential surface?
    Solution
    Equipotential surface
    $$\rightarrow $$ P.D difference between two points on the surface is zero always since potential is same everywhere in equipotential surface.
    $$\rightarrow $$ The EF is always perpendicular to the surface because there is no potential gradient along any direction parallel to the surface P so no EF parallel to the surface
    $$\rightarrow $$ Equipotential surface can have any shape not just sphere.
    $$\rightarrow $$ No work is done in moving a charge along the surface, because potential difference is zero.
    Hence option (C) is correct
  • Question 6
    1 / -0
    A parallel plate capacitor is charged. If the plates are pulled apart
    Solution
    Capacitance of a parallel plate capacitor
    C=ε0AdC=ε0Ad
    A parallel plate capacitor is charged (battery is disconnected) then the plates are pulled apart, the capacitance decreases while the charge remains the same
    Potentialdifference=ChargeCapacitance∵Potentialdifference=ChargeCapacitance
     potential difference increases
  • Question 7
    1 / -0
    What is the unit of electric potential difference?
    Solution
    Unit of electric potential difference is volt(V).
  • Question 8
    1 / -0
    Find equivalent capacitance between $$A$$ and $$B$$

    Solution
    In the given figure,
    Equivalent capcitance between $$A$$ and $$B$$.
    $${C_{eg}} = {C_1} + {C_2}$$
    $$ = 4\mu F + 4\mu F$$
    Hence,
    The equivalent capcitance between $$A$$ and $$B$$ is$$ = 8\mu F$$
  • Question 9
    1 / -0
    The net capacity across AB in  _____ $$\mu F$$.

  • Question 10
    1 / -0
    The potential of a sphere of radius $$2\ cm$$ when a charge of $$2\ coulomb$$ is given to it, will be
    Solution

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