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Electrostatic Potential and Capacitance Test - 69

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Electrostatic Potential and Capacitance Test - 69
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  • Question 1
    1 / -0

    Two identical capacitors are connected as shown in the figure. A dielectric slab is introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be :(charge on each condenser is q0q_{0}kk  = dielectric constant )

    Solution
    When capacitors are connected in series the charge present on them must be equal.
    Qc1+Qc2=2\dfrac{Q}{c_{1}}+\dfrac{Q}{c_{2}}=2 ------------ (1)
    Now when kk is introduced c2c_{2} becomes kc2kc_{2}

    Let qoq_{o} be the charge on capacitor
    given c1=c2c_{1}=c_{2}
    2qoc=2,qoc=1\Rightarrow 2\dfrac { { q }_{ o } }{ c } =2,\dfrac { { q }_{ o } }{ c } =1 from (1)
    c=qo  c ={ q }_{ o } 

    After introducing the dielectric, let charge on each plate become Q1Q_{ 1 }
    Q1qo+Q1kq0=2\Rightarrow \dfrac { Q_{ 1 } }{ { q }_{ o } } +\dfrac { Q_{ 1 } }{ k{ q }_{ 0 } } =2
    kQ1+Q1kqo=2\Rightarrow \dfrac { kQ_{ 1 }+Q_{ 1 } }{ k{ q }_{ o } } =2
    Q1=2kqok+1 \Rightarrow Q_{ 1 }=\dfrac { 2kq_{ o } }{ k+1 } 
    Q1=2qo1+1k\Rightarrow Q_{1}=\dfrac{2q_{o}}{1+ \dfrac{1}{k}}
  • Question 2
    1 / -0
    A parallel plate capacitor of capacity 5μF5 \mu F and plate separation 6cm6 cm is connected to a 1V1V battery and is charged. A dielectric of dielectric constant 44 and thickness 4cm4 cm is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is:
    Solution

  • Question 3
    1 / -0

    The capacity between the points A and B in the given circuit will be:

    Solution
    The charge distribution is as shown in the figure.
    Using Kirchhoff's law,
    Vq1/C1+(q2q1)/C3q1/C1=0V=2q1/C1(q2q1)/C3 V - q1/C_1 + (q2-q1)/C_3 - q1/C_1 =0 \Rightarrow V = 2q1/C_1 - (q2-q1)/C_3
    Vq2/C2+(q1q2)/C3q2/C2=0V=2q2/C2(q1q2)/C3 V - q2/C_2 + (q1-q2)/C_3 - q2/C_2 =0 \Rightarrow V = 2q2/C_2 - (q1-q2)/C_3
    Therefore,
    C1C3V=2C3q1C2q2+C2q1=(2C3+C1)q1C1q2 C_1C_3 V = 2C_3q1 - C_2 q2 + C_2q1 = (2C_3+C_1)q1 - C_1q2
    C2C3V=2C3q2C2q1+C2q2=(2C3+C2)q2C2q1 C_2C_3 V = 2C_3q2 - C_2q1 + C_2q2 = (2C_3 + C_2)q2 - C_2q1 
    Solving these two equations we get,

    q1=C1(C2+C3)VC1+C2+2C3q1 = \dfrac{C_1(C_2+C_3)V}{C_1+C_2+2C_3}

    q2=C2(C1+C3)VC1+C2+2C3q2 = \dfrac{C_2(C_1 + C_3)V}{C_1+C_2+2C_3}

    Equivalent capacitance =Qnet/V=(q1+q2)/V=2C1C2+C1C3+C2C3C1+C2+2C3=Q_{net}/V = (q1+q2)/V = \dfrac{2C_1C_2 + C_1C_3 + C_2C_3}{C_1 +C_2 + 2C_3}
  • Question 4
    1 / -0

    The capacitance CABC_{AB} in the given network is:

    Solution
    From the given circuit, we get  C1C2=510=12\dfrac{C_1}{C_2} = \dfrac{5}{10} = \dfrac{1}{2}
    And  C3C4=105=21\dfrac{C_3}{C_4} = \dfrac{10}{5} = \dfrac{2}{1}
    Thus  C1C2\dfrac{C_1}{C_2} is reciprocal of C3C4\dfrac{C_3}{C_4}
    For this particular type of circuit, the equivalent capacitance is given by-
    Cnet=2C1C2+C3(C1+C2)C1+C2+2C3C_{net} = \dfrac{2C_1C_2 + C_3(C_1+C_2)}{C_1+C_2+2C_3}

    Cnet=2(5)(10)+5(5+10)5+10+2(5)=17525=7μFC_{net} = \dfrac{2(5)(10) + 5(5+10)}{5+10+2(5)} = \dfrac{175}{25} = 7\mu F

  • Question 5
    1 / -0
    The equivalent capacitance CABC_{AB} of the circuit shown in the figure is :

    Solution

    The given circuit can be reconstructed as shown (As the potential difference between A and B has to same through all possible paths)

    Using the laws of combination of capacitors,

    Cnet=C1+C2+C3where C_{net} = C_1 + C_2 + C_3 \,where\,

    C2=C2,1C1=1C+1C+1(C+C2)=2C+23C=83CC1=3C8 C_2 = \dfrac{C}{2} , \dfrac{1}{C_1} = \dfrac{1}{C} + \dfrac{1}{C} + \dfrac{1}{(C+ \dfrac{C}{2})} = \dfrac{2}{C} + \dfrac{2}{3C} = \dfrac{8}{3C} \Rightarrow C_1 = \dfrac{3C}{8}

    Similarly, C3=3C8C_3 = \dfrac{3C}{8}

    Cnet=6C8+C2=5C4C_{net} = \dfrac{6C}{8} + \dfrac{C}{2} = \dfrac{5C}{4}  

  • Question 6
    1 / -0
    The effective capacity between A and B of the given network is

    Solution

  • Question 7
    1 / -0
    An air filled parallel plate capacitor has a capacitance 1pF1 pF. The separation between the plates is doubled and wax is inserted between the plates, then the capacitance becomes 2pF2 pF. What is the dielectric constant of wax ?
    Solution

    Capacitance of parallel plate capacitor with air ,Ca=ε0Ad  ....(1)C_a=\dfrac{\varepsilon _{0}A}{d}    ....(1)

    Capacitance of parallel plate capacitor with wax ,Cw=Kε0A2d .....(2)C_w=\dfrac{K\varepsilon _{0}A}{2d}  .....(2)

    (1)/(2),(1)/(2),

    CaCw=2K \dfrac{C_a}{C_w}=\dfrac{2}{K}

    12=2K\dfrac{1}{2}=\dfrac{2}{K}

    K=4\therefore K=4

  • Question 8
    1 / -0
    An isolated parallel plate capacitor is charged upto a certain potential difference. When a 3mm3mm thick slab is introduced between the plates then in order to maintain the same potential difference, the distance between the plates is increased by 2.4mm2.4mm. Find the dielectric constant of the slab. (Assume charge remains constant)
    Solution
    Cf=ε0Adt(11K)C_f=\dfrac {\varepsilon_0A}{d'-t(1-\frac {1}{K})}
    Ci=CfC_i=C_f
    ε0Ad=ε0Adt(11K)d=dt(11K)\dfrac {\varepsilon_0A}{d}=\dfrac {\varepsilon_0A}{d'-t(1-\dfrac {1}{K})}\Rightarrow d=d'-t(1-\dfrac {1}{K})
    d=d2.4×103d'=d-2.4\times 10^{ -3 }
    t=3mmt=3mm
    d=d+(24×103)3×103(11K)d=d+(2\cdot 4\times 10^{-3})-3\times 10^{-3}(1-\dfrac {1}{K})
    K=5K=5
  • Question 9
    1 / -0
    The capacity of a parallel plate condenser without any dielectric is C. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant 33, the capacity of the condenser becomes:
    Solution

    The capacitance of a parallel plate capacitor is given by

    C=εo εrAdC = \dfrac {\varepsilon_o \varepsilon_r A}{d},

    where AA is the area of the plate and dd is the distance between them.

    In the first case, there was no dielectric medium, i.e., εr=1\varepsilon_r = 1.

    So we have   C=εoAdC = \dfrac {\varepsilon_o A}{d}

    In the second case, the distance is doubled, i.e., 2d2d and a substance of dielectric constant 33 is inserted between the plates.

    Hence capacitance will change to 

    C= 3εoA2dC' = \dfrac {3 \varepsilon_o A}{2d}

    So, CC=32 \dfrac{C'}{C} = \dfrac {3}{2}

         C=32C \implies C' = \dfrac {3}{2} C

  • Question 10
    1 / -0
    A parallel plate capacitor is connected to a cell. It's positive plate A and the negative plate B have charges +Q+Q and Q-Q respectively. A third plate C identical to A and B with charge +Q+Q is  now introduced midway between the plates A and B parallel to them What is the charge on the inner surface of A ?

    Solution
    Let the capacitance of the A+B system be given by C = ϵAd\dfrac{\epsilon A}{d}
    Now when the plate is introduced in middle, the distance of A+C and C+B system is changed to d2\dfrac{d}{2} So the capacitance will be 2ϵ Ad\dfrac{2\epsilon A}{d} = 2C
    Now, Total charge in the system is Q assume A+C has q charge and C+B has Q+q charge 
    So, q2C\dfrac{q}{2C} + Q+q2C\dfrac{Q+q}{2C} = QC\dfrac{Q}{C}
    So, q = Q2\dfrac{Q}{2}

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