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Electric Charges and Fields Test - 75

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Electric Charges and Fields Test - 75
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  • Question 1
    1 / -0
    Two charges, each equal q, are kept at x=ax=-a and x=ax=a on the x-axis. A particle of mass mm and charge q0=q2{q_0} = \frac{q}{2} is placed at the origin. If  charge q0{q_0} is given a small displacement (y<<a)(y<<a) along the y-axis, the net force acting on the particle is proportional to: 
    Solution

  • Question 2
    1 / -0
    Two insulated, similar copper spheres AA and BB are separated by a distance of 0.50.5 m. Each sphere carries a charge of 6.5×107C6.5 \times {10^{ - 7}}C.
    i) What is the force of repulsion between them (F1)(F_{1})?
    ii) What will be this force if the distance between them doubled (F2)(F_{2})
    Solution

  • Question 3
    1 / -0
    Two small identical spheres each of mass 1 g1\ g and carrying same charge 109{10}^{-9} care suspended by threads of equal lengths. If the distance between the centres of the spheres is 0.3 cm0.3\ cm in equilibrium then the inclination of the thread with the vertical will be:-
    Solution

  • Question 4
    1 / -0
    A point charge q=1011 Cq=10^{-11}\ C, is placed at 4 cm4\ cm above a square plate (8 cm ×8 cm)(8\ cm\ \times 8\ cm), having charge density 0.5 ×108 C/m20.5\ \times 10^{-8}\ C/m^{2}. Find the flux related with it.
    Solution

  • Question 5
    1 / -0
    The volume charge density as a function of distance xx from one face inside a unit cube is varying as shown in the figure. Then the total flux (inS.Iunits)(in S.I units) through the cube if (p0=8085×1012Cm3)(p-{0}=8085\times10^{-12}Cm^{3}) is
  • Question 6
    1 / -0
    The electric field in a region of space is given by E=(5 i^+2j^)N/CE=(5\ \hat {i}+2\hat {j})N/C. The electric flux through an area of 2 m22\ m^{2} lying in the YZYZ plane, in S.I.S.I.. units is ?
    Solution

  • Question 7
    1 / -0
    A charge q is distributed uniformly on a ring of radius R. A sphere of equal radius R is constructed with its centre at the periphery of the ring, Find the flux of the electric field through the surface of the sphere

    Solution
    Let point of intersection of ring & sphere are P.Q.........P.Q......... let center of ring is OO & center of sphere is OO'........
    OOO O' bisects PQP Q in two equal ,if SS is the point of intersection of OOO O' & PQP Q then
    In triangle right angles POSPOS
    PO=R,OS=R/2PO=R, OS=R/2 so angle POS=θPOS=\theta
    Using trigonometry, cosθ=OS/PS=1/2 cos \theta =OS/PS=1/2
    θ=π/3\theta =\pi/3
    total angle subtended by arc is 2π/3.....2 \pi/3..... this is the arc of ring which lies inside sphere & substends
    2π/32 \pi /3 angle at the center  of ring.....
    for 2π2 \pi radian charge =q
    for unit radian =q2π = \dfrac{q}{{2\pi }}
    for q2π \dfrac{q}{{2\pi }} radian q1=(q2π)(2π3)=q3{q_1} = \left( {\dfrac{q}{{2\pi }}} \right)\left( {\dfrac{{2\pi }}{3}} \right) = \dfrac{q}{3}
    total flux through sphere= total charge enclosed q3ε0\dfrac{q}{{3{\varepsilon _0}}}
    \therefore Option CC is correct.
  • Question 8
    1 / -0
    The small spherical balls each carrying a charge Q=10μQ = 10 \mu  CC are suspended by two insulting threads of equal length 1m1m each, from a point fixed in the  cielling. It is found that is in equilibrium threads are separated by an angle 60{60^ \circ} between them, as shown in the figure. What is the tension in the threads (Given : 14πε0=9×109Nm/C2\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}Nm/{C^2})

    Solution

  • Question 9
    1 / -0
    What is the electric flux linked with closed surface?

    Solution

  • Question 10
    1 / -0
    Two point charges placed at a distance 'r' in air exert a force F. The value of distance at which they exerts same force when placed in medium (dielectric constant K) is :-
    Solution

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