Electric Charges and Fields Test

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Electric Charges and Fields Test - 61

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Question 1
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Two charges $$q$$ and $$-q$$ placed at $$(a, 0)$$ and $$(-a, 0)$$ on the x-axis. Another charge $$2q$$ is taken from $$(0,0)$$ to $$(0, a)$$, then the work done to do so

A
$$\cfrac{q}{2\pi\varepsilon_0 a}$$

B
$$\cfrac{q}{4\pi\varepsilon_0 a}$$

C
Zero

D
Infinite

Solution

When $$2q$$ is at $$(0,0)$$ $$V = 0$$ because both $$-q$$ and $$+q$$ cancel each other.
Similarly when $$2q$$ is at $$(a,0)$$ both cancel each other.
$$\begin{array}{l} { V_{ i } }=0 \\ { V_{ f } }=0 \\ \therefore \, No\, \, work\, \, is\, \, done. \end{array}$$
Hence,
option $$(C)$$ is correct answer.
Question 2
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If an electron is projected in a uniform perpendicular electric field, then its trajectory will be a

A
Straight line

B
Parabola

C
Circle

D
Helix

Solution
Question 3
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A negative charge is placed at the midpoint between two fixed equal positive charges, separated by a distance $$2d$$. If the negative charge is given a small displacement $$x (x << d)$$ perpendicular to the line joining the positive charges, how
the force (F) developed on it will approximately depend on $$x$$ ?

A
$$F\propto X$$

B
$$F\propto \dfrac{1}{x}$$

C
$$F\propto X^2$$

D
$$F\propto \dfrac{1}{x^2}$$

Solution

$$E=\dfrac{2KQ}{r^2}\cos \theta=\dfrac{2KQ}{r^2}\times \dfrac{x}{r}\therefore F=\dfrac{2KQx}{r^3}, E=\dfrac{2KQx}{(d^2+x^2)^{\dfrac{3}{2}}}\therefore =\dfrac{-2KQqx}{(d^2+x^2)^{\dfrac{3}{2}}}$$
for $$x<<d, x^2$$ can be neglected $$\therefore =\dfrac{-2KQqx}{d^3}, F\propto x$$
Question 4
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A charged oil drop is suspended in a uniform field of 3 x 10⁴ v/m so that it neither falls nor rises. The charge on the drop will (Take the mass of the charge = 9.9 x 10^-15 kg and g = 10 m/s² )

A
$$1.25\times 10^{10} C$$

B
$$2.5\times 10^{10} C$$

C
$$3.3 \times {10^{ - 18}}C$$

D
none

Solution

$$\begin{array}{l} At\, \, equilibrium,\, \, electric\, \, force\, \, on\, \, drop\, \, balances\, \, weight\, \, of\, \, drop. \\ qE=mg \\ q=\dfrac { { mg } }{ E } \\ q=\dfrac { { 9.9\times { { 10 }^{ -15 } } } }{ { 3\times { { 10 }^{ 4 } } } } =3.3\times { 10^{ -18 } }C \end{array}$$
Hence,
option $$(C)$$ is correct answer.
Question 5
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In Region of Electric field Given by $$\vec{E} = (Ax + B) \hat{i}$$. Where $$A = 20$$ unit and $$B = 10$$ unit. If Electric potential at $$x = 1\,m$$ is $$V_1$$ and at $$x = -5 \,m$$ is $$V_2$$. Then $$V_1 - V_2$$ is equal to

A
$$150 \,V$$

B
$$170 \,V$$

C
$$140 \,V$$

D
$$180 \,V$$

Solution

$$v_f - v_i = -\int E.dx$$

$$v_1 - v_2 = \displaystyle \overset{1}{\underset{-5}{\int}} (Ax + B)dx$$

$$= -\left(\dfrac{Ax^2}{2} + Bx\right)^1$$

$$= -\left[\left(\dfrac{A}{2} + B\right) - \left(\dfrac{25A}{2} - 5B\right)\right]$$

$$= - \left[\dfrac{20}{2} + 10 - 250 +50\right]$$

$$= [10 + 10 + 50 - 250]$$
$$= + 180$$
Question 6
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Two positive charges and an negative charge, all of equal magnitude, are set at the corners of an equilateral triangle.
Which diagram best represents the electric field surrounding the charges?

A

B

C

D

Solution

Option A is correct. Electric field line start from positive charges and terminate at negative charge, and lines from same type charge don't intersect or meet.
Question 7
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Two masses $$M_1$$ and $$M_2$$ carry positive charges $$Q_1$$ and $$Q_2$$, respectively. They are dropped to the floor in a laboratory setup from the same height where there is a constant electric field vertically upwards $$M_1$$ hits the floor before $$M_2$$. Then :

A
$$Q_1 > Q_2$$

B
$$Q_1 < Q_2$$

C
$$M_1Q_1 > M_2Q_2$$

D
$$M_1 Q_2 > M_2Q_1$$

Solution

$$Q_1=\dfrac{M_1g-Q_1E}{M_1},Q_2=\dfrac{M_2g-Q_2E}{M_2}$$
as $$S=ut+\dfrac{1}{2}at^2$$
So, $$h=a\times t_1+\dfrac{1}{2}(g-\dfrac{Q_1E}{M_1})t^2_1$$.........(1)

$$h=a\times t_2+\dfrac{1}{2}(g-\dfrac{Q_2E}{M_2})t^2_2$$........(2)
given $$t_1 < t_2 \Rightarrow t^2_1 < t^2_2$$

So, $$\dfrac{2h}{g-\dfrac{Q_1E}{M_1}} < \dfrac{2h}{g-\dfrac{Q_2E}{M_2}}$$

$$g-\dfrac{Q_1E}{M_1} > g-\dfrac{Q_2E}{M_2}$$

$$\dfrac{Q_1E}{M_1} < \dfrac{Q_2E}{M_2}$$

$$\Rightarrow \boxed{Q_2M_1 > Q_1M_2}$$
Question 8
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An electron passes into the space between two parallel plates that are 5.0 cm apart and which
are maintained at electric potentials of +2000V and -500V, respectively.
What is the electric force on the electron?

A
$$1.6 \times 10^{-15}N$$

B
$$4.8 \times 10^{-15}N$$

C
$$6.4 \times 10^{-15}N$$

D
$$8.0 \times 10^{-15}N$$

Solution

We know that
$$E = \dfrac{V}{l}$$
$$\implies E = \dfrac{{2000-(-500)}}{0.05}\ Vm^{-1}$$
$$\implies E = 50000\ Vm^{-1}$$
So, electric force $$F= eE$$
$$\implies F = 1.6\times 10^{-19}\times 50000 N$$
$$\implies F = 8.0\times 10^{-15}\ N$$
Question 9
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One brass plate is inserted between two charges. The force between two charges will

A
Remain the same

B
Increase

C
Decrease

D
Fluctuate

Solution
Question 10
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Three point charges $$4q,Q$$ and $$q$$ are placed in a straight line of length$$L$$ at points $$0,\dfrac{L}{2}$$ and $$L$$ respectively. The net force on charge $$q$$ is zero. The value of $$Q$$ is

A
$$4q$$

B
$$-q$$

C
$$-0.5q$$

D
$$-2q$$

E
$$3q$$

Solution

Net force on charge, $$q=0$$
$$\Rightarrow f_1+f_2=0$$..........(i)
Here, $$f_1$$ is the force on charge $$q$$ due to charge $$4q$$.
$$f_2$$ is the force on charge q due to charge $$Q$$.
$$\therefore$$ From equation (i),
$$\Rightarrow \dfrac{K(4q)q}{L^2}+\dfrac{KQq}{(\dfrac{L}{2})^2}=0$$
$$\Rightarrow 4q^2+4Qq=0$$
$$\Rightarrow Q=-q$$