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Electric Charges and Fields Test - 22

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Electric Charges and Fields Test - 22
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  • Question 1
    1 / -0
    A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre 
    Solution
    Charge Q will be distributed over the surface of hollow metal sphere. 
    (i) For r < R (inside)
    By Gauss law, $$\oint \bar{E_{in}} . \overline{dS} = \dfrac{q_{en}}{\varepsilon_0}= 0$$
    $$\Rightarrow E_{in} = 0$$            $$(\because q_{en} = 0)$$
    (ii) For r > R (out side)
    $$\oint \bar{E}_0 . \overline{dS} = \dfrac{q_{en}}{\varepsilon_0}$$
    Here, $$q_{en} = Q (\because q_{en} = Q)$$
    $$\therefore E_0 4 \pi r^2 = \dfrac{Q}{\varepsilon _0}$$
    $$\therefore E_0 \propto \dfrac{1}{r^2}$$

  • Question 2
    1 / -0
    Electric field outside a long wire carrying charge q is proportional to :
    Solution
    Using Gauss Law across the cylindrical Gaussian surface,
    $$\displaystyle\int E.dS = \dfrac{q_{enc}}{\epsilon_0}$$
    $$\therefore E\times 2\pi r \times l = \dfrac{\lambda\times l}{\epsilon_0}$$
    $$\therefore E = \dfrac{\lambda}{2\pi r\epsilon_0}$$
    Here $$\lambda$$ is the linear charge density on the wire.

  • Question 3
    1 / -0
    An electric line of force in the $$xy-plane$$ is given by equation $$x^{2} + y^{2} = 1$$. A particle with unit positive charge, initially at rest at the point $$x = 1, y = 0$$ in the $$xy - plane$$.
    Solution
    Charge will move along the circular line of force because $$x^{2} + y^{2} = 1$$ is the equation of circle in $$xy - plane$$.
  • Question 4
    1 / -0
    Two identical charged spheres suspended from a common distance $$d(d < < 1)$$ apart because of their mutual repulsion. The charged begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity $$v$$. Then $$v$$ varies as a function of the distance $$x$$ between the spheres, as
    Solution
    From figure $$\tan \theta = \dfrac {F_{e}}{mg} =\theta$$
    $$\dfrac {kq^{2}}{x^{2}mg} = \dfrac {x}{2l}$$
    or $$x^{3}\propto q^{2} .... (1)$$
    or $$x^{3/2} \propto q ..... (2)$$
    Differentiating eq. $$(1)$$ w.r.t time
    $$3x^{2} \dfrac {dx}{dt}\propto 2q \dfrac {dq}{dt}$$ but $$\dfrac {dq}{dt}$$ is constant
    so $$x^{2}(v) \propto q$$ Replace $$q$$ from eq. $$(2)$$
    $$x^{2}(v) \propto x^{3/2}$$ or $$v\propto x^{-1/2}$$.
  • Question 5
    1 / -0

    Two unlike charges attract each other with a force of 10N. If the distance between them is doubled, the force between them is :

    Solution
    We know $$F=\dfrac {1}{4\pi \varepsilon_0}\dfrac {q_1q_2}{r^2}=10 N$$

    if distance is doubled.
    $$F_{new}=\dfrac {1}{4\pi \varepsilon_0}\cdot \dfrac {q_1q_2}{(2r)^2}=\dfrac {10}{4}$$

    $$=2\cdot 5 N$$
  • Question 6
    1 / -0

    Number of electric lines of force emanating from $$1C$$ of positive charge in vacuum is :

    Solution
    $$\phi $$ from $$1C $$ of charge is $$= \dfrac{Q_{enclosed}}{\epsilon_o}$$
    $$= \dfrac{1}{8.8501\times 10^{-12}}$$

    $$= 1.129\times 10^{11}$$
  • Question 7
    1 / -0

    A force ‘F’ is acting between two charges in air. If the space between them be completely filled with a medium $$K=4$$, the force will be :

    Solution
    By Coulomb's law, the force between two charges $$(q_1, q_2)$$ in air is given by $$F=\dfrac{q_1q_2}{4\pi\epsilon_0 r^2}$$  
    where $$r=$$ distance between two charges.
    Now the force between two charges $$(q_1, q_2)$$ in medium with dielectric constant $$K=4$$ becomes, $$F'=\dfrac{q_1q_2}{4\pi K\epsilon_0 r^2}$$
    So, $$F'=\dfrac{F}{K}=\dfrac{F}{4}$$
  • Question 8
    1 / -0

    An electron revolves around the nucleus of hydrogen atom in a circle of radius $$5\times 10^{-11}\ m$$The intensity of electric field at a point in the orbit of the electron is:

    Solution
    We know hydrogen atom consist of one proton and one electron
    $$R=5\times 10^{-11} m$$
    We know attractive force $$=\dfrac {k q_1(q_2)}{r^2}=\dfrac {9\times 10^9\times (1\cdot 6\times 10^{-19})^2}{(5\times 10^{-11})^2}$$
    We know   $$F = E q$$
    $$\therefore E=\dfrac {F}{q}$$
    $$\therefore E=\dfrac {9\times 10^9\times (1\cdot 6\times 10^{-19})}{5\times 10^{-11})^2}=5\cdot 76\times 10^{11} N/C.$$

  • Question 9
    1 / -0
    A charge $$q_1$$ exerts a force of $$45N$$ on a charge of $$q_{2}=10^{-5} C$$ located at a point $$0.2m$$ from $$q_{1}$$. The magnitude of $$q_{1}$$ is
    Solution
    As we know electrostatic force between two charges $$F=\dfrac {k q_1 q_2}{r^2}$$

    $$\Rightarrow 45=\dfrac {(9\times 10^9)(q_1)(10^{-5})}{(0\cdot 2)^2}$$

    $$\Rightarrow q_1=\dfrac {45\times 0\cdot 2\times 0\cdot 2}{9\times 10^9\times 10^{-5}}$$

    $$\Rightarrow q_1=2\times 10^{-5}C$$
  • Question 10
    1 / -0
    When we remove polyester or woollen clothes in dark, we can see a spark and hear a crackling sound. Which of the following is responsible for it?
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