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Electric Charges and Fields Test - 29

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Electric Charges and Fields Test - 29
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  • Question 1
    1 / -0
    In the diagram, three point charges (labeled 1, 2 and 3) are shown, along with the electric field around them. Which charge(s) is/are positive?

    Solution
    The electric field always originate at the positive charge and terminate at the negative charge.
    From figure, the electric field lines originate at the points 1 and 3 but ends at point 2.
    Thus points 1 and 3 represent positive charge and 2 represents the negative charge.
  • Question 2
    1 / -0
    An electron and a proton are both released from rest 1 meter from a large stationary negative charge, considering only the force from the large stationary negative charge on the proton and electron. Which of the following is true?
    Solution
    The magnitude of electric force between electron and proton is given by $$F=\dfrac{1}{4\pi\epsilon_0}\dfrac{e(-e)}{d^2}$$
    The force on each of them is equal in magnitude and is attractive.
    The acceleration of a species is given by $$a=\dfrac{F}{m}$$
    Since the mass of proton is larger, the acceleration of electron is greater.
  • Question 3
    1 / -0
    Electric lines of force about a positive point charge are:
    Solution
    Electric lines of force about a positive point charge are always radially outwards.

  • Question 4
    1 / -0
    Electric charge is measured in
    Solution
    The standard unit of charge is coulomb (C). Here ampere is the unit of current, volts is the unit of electric potential difference and watts is the unit of power. 
  • Question 5
    1 / -0
    The charge is negative, then the electric lines of forces are 
    Solution
    Lines of force originate from a positive charge and terminate to a negative charge. Hence, when the charge is negative, then the electric lines  of force are straight lines converging towards the charge.
  • Question 6
    1 / -0
    The force acting between two point charges kept at a certain distance is $$5$$N. Now the magnitudes of charges are doubled and distance between them is halved, the force acting between them is _________ N.
    Solution
    According to Coulomb's law, force between two charge particle $$q_{1}$$ and $$q_{2}$$ at a distance r is $$\dfrac{Kq_{1}q_{2}}{r^{2}}$$.
    Given, Initial Force = 5N =  $$\dfrac{Kq_{1}q_{2}}{r^{2}}$$.
    Now, $$Q_{1}$$ = 2$$q_{1}$$  and $$Q_{2}$$ = 2$$q_{2}$$  and R= $$ \dfrac{r}{2} $$. Now , Force = $$\dfrac{KQ_{1}Q_{2}}{R^{2}}$$ =  $$\dfrac{ 2\times 2\times K2q_{1}2q_{2}}{r^{2}}$$ =  $$ 16\times \dfrac{Kq_{1}q_{2}}{r^{2}}$$ = $$16\times 5N$$ = 80N.
    Therefore, D is the correct option.
  • Question 7
    1 / -0
    Coulomb's law relates two charges and distance between them describing the electric force as being:
    Solution
    Coulomb's law can be given by:
    $$F = k\cfrac{q_{1} q_{2} }{r^{2}}$$
    $$\therefore F \propto \cfrac{1}{r^{2}}$$
  • Question 8
    1 / -0
    Electric lines of force about negative point charge are
    Solution
    Correct Answer: Option (D)
    Around a negative charge force converge and moves inside, so it’s radial and inward.
  • Question 9
    1 / -0
    The electric field inside a conductor.
    Solution

  • Question 10
    1 / -0
    The force between two charges $$2\mu C$$ and $$4\mu C$$ is $$24 N$$, when they are separated by a certain distance in free space. The forces if, $$(a)$$ distance between them is doubled and $$(b)$$ distance is halved are :
    Solution
    We know $$F=\dfrac {kq_1q_2}{r^2}=24N$$
    a) When $$r$$ is doubled

    $$F=\dfrac {kq_1q_2}{(2r)^2}=\dfrac {kq_1q_2}{4r^2}=\dfrac {24}{4}=6N$$

    b) When $$r$$ is halved

    $$F=\dfrac {kq_1q_2}{\bigg (\dfrac {r}{2}\bigg )^2}=\dfrac {4kq_1q_2}{r^2}=4\times 24=96N$$
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