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Electric Charges and Fields Test - 30

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Electric Charges and Fields Test - 30
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  • Question 1
    1 / -0

    A charge of 1010 unitsunits is divided so that force between the two charges is maximum when placed 22 cmcm apart. The two charges are:

    Solution
    Let the two charges be q1,q2q_1, q_2
    q1+q2=10q_1+q_2=10
    Force between them when placed 2 cm apart
    F=k(q1)(q2)(2×102)2F=\dfrac {k(q_1)(q_2)}{(2\times 10^{-2})^2}
    F=k(q1)(10q1)(2×102)2F=\dfrac {k(q_1)(10-q_1)}{(2\times 10^{-2})^2}
    differentiate FF and equate it to zero to find at value of q1q_1, FF will be maximum.
    dFdq1=k(2×102)2(102q1)=0\dfrac {dF}{dq_1}=\dfrac {k}{(2\times 10^{-2})^2}(10-2q_1)=0
    q1=5\Rightarrow q_1=5
    q2=10q1q_2=10-q_1
    q2=5q_2=5
    q1=q2=5 C\therefore q_1=q_2=5\ C
  • Question 2
    1 / -0

    The electric field in a region of space is given by E=(5i^+2j^ )Nc1\vec{E}=(\hat{5i}+\hat{2j}  )Nc^{-1} . The electric flux due to this field through an area 2m22m^{2} lying in the Y-Z plane in S.I. units is

    Solution
    As 2m22m^{2} area is in y-z plane its area vector which is perpendicular to y-z plane is in x axis.
    \therefore area vector =2(i^)=2(\hat{i})
    given E=5i+2jE=5i+2j
    We know flux ϕ=E.A\phi=E.A
    =(2i).(5i+2j)=(2i).(5i+2j)
    ϕ=10\phi=10
  • Question 3
    1 / -0
    A charge QQ is divided into two parts q1q_{1} and q2q_{2} such that they experience maximum force of repulsion when separated by certain distance. The ratio of Q, q1Q, q_{1} and q2q_{2} is :
    Solution
    Step 1: Divide Q\textbf{Step 1: Divide Q}
    QQ is divided into two parts q1q_1 andq2q_2 
                \therefore      Q=q1+q2Q = q_{1} + q_{2}                                                                                                                                        ....(1)....(1)

    Step 2: Force of repulsion between q1 and q2\textbf{Step 2: Force of repulsion between}\ q_{1}\ \textbf{and}\ q_{2} 
    Let rr be the distance between charges. 
     Using Coulomb's law,  F=kq1q2r2F = \dfrac{k q_{1} q_{2}}{r^2}   =kq1(Qq1)r2   = \dfrac{k q_{1} (Q - q_{1})}{r^2}                                                                                          ....(2)....(2)

    Step 3: Condition for force to be maximum\textbf{Step 3: Condition for force to be maximum}  
    The repulsive force between the charges will be maximum if,
                          dFdq1=0\dfrac{dF}{dq_{1}} = 0
                      kr2(Q2q1)=0\Rightarrow\dfrac{k}{r^2} (Q - 2q_{1}) = 0
                      q1=Q2\Rightarrow q_{1} = \dfrac{Q}{2}

     Equation (1)(1)\Rightarrow   q2=QQ2=Q2q_{2} = Q - \dfrac{Q}{2} = \dfrac{Q}{2}  

    Step 4: Ratio of Q:q1:q2\textbf{Step 4: Ratio of}\ Q : q_{1} : q_{2}   

    Q:q1:q2=Q:Q2:Q2=1:12:12Q : q_{1} : q_{2} = Q : \dfrac{Q}{2} : \dfrac{Q}{2} = 1 : \dfrac{1}{2} : \dfrac{1}{2}  
    Q:q1:q2=2:1:1\Rightarrow Q : q_{1} : q_{2} = 2 : 1 : 1

    Hence, Option (D) is correct.
  • Question 4
    1 / -0

    Two alpha particles are separated by a distance of 1013m10^{-13}m. The force between them in free space is :

    Solution
    α  particles  are He+2 ions\alpha  \ particles \  are\ He^{+2} \ ions
    \therefore Net charge present on them is 2×(16×1019)C.2\times (1\cdot 6\times 10^ {-19})C.
    We know F=kq1q2r2F=\dfrac {kq_1q_2}{r^2}

    F=(9×109)(2×16×1019)(2×16×1019)(1013)2F=\dfrac {(9\times 10^9)(2\times 1\cdot 6\times 10^{-19})(2\times 1\cdot 6\times 10^{-19})}{(10^{-13})^2}

    F=9216×10291026F=\dfrac {92\cdot 16\times 10^{-29}}{10^{-26}}

    F=9216×103NF=92\cdot 16\times 10^{-3}N
  • Question 5
    1 / -0

    The excess (equal in number) number of electrons that must be placed on each of the two small spheres spaced 3cm3 cm apart with a force of repulsion between them to be 1019N10^{-19} N is :

    Solution
    We need force of repulsion =1019N= 10^{-19}N
    Let n no. of increase electrons are placed on each sphere 
    K(ne)(ne)(3×102)2=1019\Rightarrow \dfrac{K(ne)(ne)}{(3\times 10^{-2})^2}= 10^{-19}

    n2e2=1019×9×1049×109×e2\Rightarrow n^2e^2= \dfrac{10^{-19}\times 9\times 10^{-4}}{9\times 10^9\times e^2}

    n2=1032e2n=1016e\Rightarrow n^2= \dfrac{10^{-32}}{e^2} \Rightarrow n= \dfrac{10^{-16}}{e}

    n=10161.6×1019n= \dfrac{10^{-16}}{1.6\times 10^{-19}}

    n= 10001.6=625n=  \dfrac{1000}{1.6}= 625
  • Question 6
    1 / -0

    Two charges each of 100100 micro coulomb are separated in a medium of relative permittivity 22 by a distance of 5cm5 cm. The force between them is :

    Solution
    We know force between two charges in a medium separated by a distance 'r'
    F=(q1)(q2)4πεr2F=\dfrac {(q_1)(q_2)}{4\pi \varepsilon r^2}

    where, ϵr=ϵϵo \epsilon_r= \dfrac{\epsilon}{\epsilon_o}
    given, ϵr=2\epsilon_r=2     ϵ=2×ϵo\implies \epsilon = 2\times \epsilon_o

    F=(100×106)(100×106)×9×1092×(5×102)2F=\dfrac {(100\times 10^{-6})(100\times 10^{-6})\times 9\times 10^9}{2\times (5\times 10^{-2})^2}

    F=902×25×104F=\dfrac {90}{2\times 25\times 10^{-4}}

    F=90×100×10250F=\dfrac {90\times 100\times 10^2}{50}

    F=18×104NF=1\cdot 8\times 10^4 N
  • Question 7
    1 / -0
    Two equally charged pith balls 3 cm3\ cm apart repel each other with a force of 4×105 N4\times10^{-5}\ N . The charge on each ball is
    Solution
    The pith balls repel each other with force F=4×105 NF=4\times 10^{-5}\ N and let the charge on each pith ball be QQ.

    k(Q)(Q)(3×102)2=4×105\therefore \dfrac {k(Q)(Q)}{(3\times 10^{-2})^2}=4\times 10^{-5}

    or, Q2=4×105×9×1049×109Q^2=\dfrac {4\times 10^{-5}\times 9\times 10^{-4}}{9\times 10^9}

    or, Q2=4×1018Q^2=4\times 10^{-18}

    or, Q=2×109 CQ=2\times 10^{-9}\ C

  • Question 8
    1 / -0

    Q1,Q2Q_{1},Q_{2} charges are separated by distance d‘d’. FF is the force between them. Now the value of Q2Q_{2} is halved. To have the same force between the charges, the distance of separation should be :

    Solution
    Initially force between them F=KQ1Q2d2F= \dfrac{K Q_1 Q_2}{d^2}
    Now Q2Q_2 is halved so to have the same force let the distance be X

    K(Q22)Q1X2=KQ1Q2d2\Rightarrow \dfrac{K(\dfrac{Q_2}{2})Q_1}{X^2}= \dfrac{K Q_1 Q_2}{d^2}

    12X2=1d2\Rightarrow \dfrac{1}{2X^2}= \dfrac{1}{d^2}

    12X=1d\Rightarrow \dfrac{1}{\sqrt2X}= \dfrac{1}{d}

    X=d2\Rightarrow X= \dfrac{d}{\sqrt2}
  • Question 9
    1 / -0

    Two unlike charges separated by a distance of 1m1m attract each other with a force of 0.108N0.108N. If the charges are in the ratio 1:31:3, the weak charge is :

    Solution
    Let the charges be qq and 3q3q
    given force =0.108N= 0.108N
    0.108=K(q)(3q )(1)2\Rightarrow 0.108= \dfrac{K(q )(3q )}{(1)^2}
    0.108=9×109×3q2\Rightarrow 0.108 = 9\times 10^9\times 3q ^2

     q2=108×1039×3×109\Rightarrow q ^2= \dfrac{108\times 10^{-3}}{9\times 3\times 10^9}

    q2=4×1012\Rightarrow q ^2= 4\times 10^{-12}
    q=2×106\Rightarrow q = 2\times 10^{-6}
    q=2μC\therefore q = 2 \mu C
  • Question 10
    1 / -0

    Two charges 2C2C and 6C6C are separated by a finite distance. If a charge of 4C-4C is added to each of them, The initial force of 12×103N12 \times10^{3}N will change to :

    Solution
    Given initial force =12×103=12\times 10^3
    k(2)(6)x2=12×103\Rightarrow \dfrac{k(2)(6)}{x^2}=12\times 10^3
    when a charge of 4c- 4c is added to each of them
    Fnew=k(2)(2)x2F_{new}=\dfrac{k(-2)(2)}{x^2}
    (Putting value of x2x^2 from above equation)

    =k(2)(2)k(2)(6)×12×103=\dfrac{k(-2)(2)}{k(2)(6)}\times{12}\times 10^3

    Fnew=4×103NF_{new}=-4\times 10^3N (negative sign implies attraction)
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