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Electric Charges and Fields Test - 41

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Electric Charges and Fields Test - 41
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  • Question 1
    1 / -0
    A positive charged particle is kept on the plane surface. There are two points X and Y lies on the plane surface as shown above. If the point Y is three times as far away from Q as point X. What is the ratio of the electric force that would act on a small charge placed at point Y compared to the charge placed at point X?

    Solution
    Force on a charged particle at a distance $$d$$ from the other charge=$$\dfrac{1}{4\pi\epsilon_0}\dfrac{Qq}{d^2}$$.
    Thus the force at X=$$F_X=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qq}{d^2}$$
    Force at Y=$$F_Y=\dfrac{1}{4\pi\epsilon_0}\dfrac{Qq}{(3d)^2}$$
    Thus the force at X is $$9$$ times than at Y.
  • Question 2
    1 / -0
    On decreasing the charge on both particles by a factor of $$3$$ while leaving all other factors the same will:
    Solution
    Let the initial charges of both the particles be  $$q_1$$ and  $$q_2$$.
    Thus initial coulomb force       $$F = \dfrac{kq_1q_2}{r^2}$$
    Now the charge of both the particles is reduces by a factor of 3,  i.e   $$q'_1 = \dfrac{q_1}{3}$$  and   $$q'_2 = \dfrac{q_2}{3}$$
    Thus new coulomb force between them       $$F = \dfrac{k (q_1/3) \times (q_2/3)}{r^2}  =\dfrac{F}{9}$$
    Hence the force decreases by a factor of  $$9$$.
  • Question 3
    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 electric force acting on a charged species at a finite distance is $$\dfrac{1}{4\pi\epsilon_0}\dfrac{Qq}{d^2}$$
    Since charge of both proton and electron is same but opposite in sign, the magnitude of force on both is same, but direction is opposite.
  • Question 4
    1 / -0
    A particle of charge +2q exerts a force F on a particle of charge -q . What is the force exerted by the particle of charge -q on the particle of charge +2q?
    Solution
    The force exerted by a charged particle $$q_1$$ on another charged particle $$q_2$$ is $$F_{12} = \displaystyle \frac{q_1q_2}{4\pi\epsilon_0r^3}\vec{r}$$
    Here, $$\vec{r} $$ is the position vector from $$q_1$$ to $$q_2$$
    Thus, the force exerted by $$q_2$$ on $$q_1$$ is $$F_{21} = \displaystyle \frac{q_1q_2}{4\pi\epsilon_0r'^3}\vec{r'}$$
    We know that $$\vec{r'} = -\vec{r}$$
    Thus, $$\Rightarrow r' = |\vec{r'}| = |-\vec{r}| = r$$
    Thus, $$F_{21} = -F_{12}$$
  • Question 5
    1 / -0
    Two charged particle of charge $$+2q$$ & $$-q$$ are kept at some distance .A particle of charge $$+2q$$ exerts a force $$F$$ on a particle of charge $$-q$$. Find out the force exerted by the particle of charge $$-q$$ on the particle of charge $$+2q$$?
    Solution
    As the given charges are opposite ,there will  attract each other.it means charge $$+2q$$  pulls the charge $$-q$$ towards itself and same activity will be done by charge  $$-q$$ ,they exert force on each other in opposite direction direction.
         now , electrostatic force between them 
                $$F=\dfrac{1}{4\pi\varepsilon}\dfrac{2q\times q}{r^{2}}$$
      that is the force exerted by one charge on another therefore $$-q$$ will exert equal force on charge $$+2q$$ but in opposite direction  i.e  $$-F$$
  • Question 6
    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 electron has negative charge ,$$-e$$ and proton has positive charge ,$$+e$$.
    Thus, the force on electron due to the large stationary charge will be repulsive and force on proton due to the large stationary charge will be attractive.
    Since both electron and proton released from same distance and they have equal magnitude of charge so the magnitude of force on both the electron and proton will be same. 
    Thus, option C will be correct.  
  • Question 7
    1 / -0
    Two charged particles exert a force of magnitude F on one another. If the distance between them is doubled and the charge of one of the particles is doubled, what is the new force acting between them?
    Solution
    By Coulomb's law, $$F=\dfrac{kq_1q_2}{r^2}$$
    If the distance between them is doubled and the charge of one of the particles is doubled, the force becomes, $$F'=\dfrac{kq_1(2q_2)}{(2r)^2}=F/2$$
  • Question 8
    1 / -0
    Two charged particles exert a force of magnitude $$F$$ on one another. If the distance between them is doubled and the charge of one of the particles becomes four times, Find out the new force acting between them?
    Solution
    Let the charge of the particles be  $$Q_1 $$ and $$Q_2$$ and distance between them be $$r$$.
    Force between them       $$F = \dfrac{K Q_1Q_2}{r^2}$$

    Now    $$Q_1' = 4Q_1$$  and   $$r' =2r$$
    New force between them       $$F' = \dfrac{K Q_1' Q_2}{(r')^2}  = \dfrac{K (4Q_1) Q_2}{4r^2}  = F$$
  • Question 9
    1 / -0
    Consider two charges located on a line in a region far from other charges. Charge 1 is located at position x = 0 m on the line, and Charge 2 is located at position x = 12 m on the line.
    A proton placed at any point on the line between the two charges will experience a net force due to the two charges. The net force acting on the proton is shown as a function of position in the graph, where positive values of force represent rightward force.
    What is the sign of each charge, and which charge is stronger?

    Solution
    The Coulomb force experienced by a charged particle is $$\dfrac{1}{4\pi\epsilon_0}\dfrac{Qq}{d^2}$$.
    where $$q,Q$$ are the respective charges and $$d$$ is the separation between them.
    Since force is positive in the left part of the curve, the charge placed at $$x=0m$$ is positive. Since the force is negative in right half of the curve, the charge placed at $$x=12m$$ is negative.
    The net force at a point is resultant of the two forces at that point. Since at the exact mid-point where $$d$$ of both the curves is same(at $$x=6m$$), the net force at that point is positive.
    Hence $$\left|Q_+\right|>\left|Q_-\right|$$
  • Question 10
    1 / -0
    A charge $$Q$$, far from other charges, is fixed a distance $$\cfrac{1}{2}s$$ above the center of a square with side length s as shown in the diagram. What is the value of the electric flux that passes through the square due to the charge $$Q$$?

    Solution
    Let the Gaussian cube of side $$s$$ such that the charge is placed at its centre.
    Thus electric flux passing through the cube      $$\phi_{cube} = \dfrac{Q}{\epsilon_o}$$
    As the charge is placed symmetric about all the 6 faces of cube, thus flux passing through one face is equal to the one-sixth of the total flux passing through the entire cube.
    $$\therefore$$ Flux passing through the square        $$\phi_1 = \dfrac{\phi_{cube}}{6} = \dfrac{Q}{6\epsilon_o}$$
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