Electric Charges and Fields Test

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Electric Charges and Fields Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

Given that $$q_1 + q_2 = q$$. If the force between $$q_1$$ and $$q_2$$ is maximum, then $$\dfrac{q_1}{q_2}$$ is

A
$$1$$

B
$$0.75$$

C
$$0.5$$

D
$$0.25$$

Solution

$$F=\dfrac{Kq_1q_2}{r^2}$$

$$\implies F=\dfrac{Kq_1(q-q_1)}{r^2}$$

Differentiating both sides with respect to $$q_1-$$

$$\dfrac{K}{r^2}\left(q-2q_1\right)=\dfrac{dF}{dq_1}$$

For, force to be maximum, $$\dfrac{dF}{dq_1}=0$$

$$\implies q-2q_1=0$$

$$\implies q_1=\dfrac{q}{2}$$

$$q_2=q-q_1=\dfrac{q}{2}$$

Hence, $$\dfrac{q_1}{q_2}=1$$

Answer-(A)
Question 2
1 / -0

Two small balls, each carrying a charge q are suspended by equal insulator string of length $$\dfrac{l}{2}$$ m from the hook of a stand. This arrangement is carried in a satellite in space. The tension in each string will be :

A
$$\dfrac{1}{4 \pi \varepsilon_0} \dfrac{q}{l^2}$$

B
$$\dfrac{1}{4 \pi \varepsilon_0} \dfrac{q^2}{4l^2}$$

C
$$\dfrac{1}{4 \pi \varepsilon_0} \dfrac{q^2}{l^2}$$

D
$$\dfrac{1}{(4 \pi \varepsilon_0)} \dfrac{q}{l}$$

Solution

As we know that there is no gravity in space. Thus no gravitational force acts on the ball in the downward direction.
In space, the only forces acting on a charge are the tension of the spring (upwards) and the electrostatic force which are balanced.
Thus, tension in the string $$T=\dfrac { 1 }{ 4\pi { \epsilon }_{ 0 } } \dfrac { { q }^{ 2 } }{ { l }^{ 2 } } $$
Question 3
1 / -0

The concept of solid angle is a natural extension of a angle to dimensions.

A
plane, two

B
line, three

C
plane, three

D
line, two

Solution

The concept of solid angle is a natural extension of two-dimensional plane angle to three-dimensional. It is measured in terms of the unit called steradian, abbreviated as sr.
Question 4
1 / -0

Calculate the solid angle subtended by an octant of a sphere at the centre of the sphere.

A
$$\dfrac{\pi}{4}$$

B
$$\dfrac{\pi}{3}$$

C
$$\dfrac{\pi}{2}$$

D
$${\pi}$$

Solution

An octant of sphere is each part of sphere divided by a 3-axis cartesian system plane.

Its surface area is $$S=\dfrac{1}{8}\times 4\pi r^2=\dfrac{\pi r^2}{2}$$

Solid angle is given by $$\Omega=\dfrac{S}{r^2}$$

$$=\dfrac{\pi}{2}$$

Answer-(C)
Question 5
1 / -0

A short electric dipole (which consists of two point charges $$+q \,\, and -q$$) is placed at the centre O and inside a large cube (ABCDEFGH) of length L, as shown in figure. The electric flux, emanating through the cube is:

A
$$q/4 \pi \varepsilon _0L$$

B
Zero

C
$$q/2 \pi \varepsilon _0L$$

D
$$q/3 \pi \varepsilon _0L$$

Solution

Total charge enclosed inside the cube $$= 0$$. SO, the electric flux$$=\dfrac{q}{\epsilon_0}=0$$
Question 6
1 / -0

Directions For Questions

A hollow cylindrical box of length $$1m$$ and area of cross-section $$25cm^2$$ is placed in a 3D coordinate system as shown. The electric field in the region is given by $$E=50x \hat i$$ where $$E$$ is in $$N/C$$ and $$x$$ is in $$m$$.

...view full instructions

Find the net flux through the cylinder.

A
$$-0.125Nm^2/C$$

B
$$-0.25Nm^2/C$$

C
$$0.25Nm^2/C$$

D
$$0.125Nm^2/C$$

Solution

Flux through the surface $$\phi=\int \overrightarrow{E}.\overrightarrow{dS}$$

Electric field $$\overrightarrow{E}$$ is constant.
So, flux $$=E\times Area$$
$$=50\times 25\times 10^{-4}$$
$$=1250\times 10^{-4}$$
$$=0.125\: NC^{-1}m^{2}$$
Question 7
1 / -0

What must be the distance between two equal and opposite point charges (say $$+q$$ and $$-q$$) for the electrostatic force between them to have a magnitude of $$16\ N$$?

A
$$4\sqrt {kg}\ metre$$

B
$$\dfrac {q}{4}\sqrt {k}\ metre$$

C
$$4 kq\ meter$$

D
$$\dfrac {4k}{q}\ metre$$

Solution

Electrostatic force between two equal and opposite charges is $$16\ N$$.
$$F = \dfrac{k \space q(-q)}{r^2}$$

$$\left | \dfrac{- k \space q^2}{r^2}\right |= 16\ N$$

$$r = \sqrt{\dfrac{k \space q^2}{16}} $$

$$r=\dfrac{q}{4} \sqrt k$$ meters
Question 8
1 / -0

Electric flux per unit solid angle is defined as

A
Electric force

B
Electric field intensity

C
Electric potential

D
Electric power

Solution

Electric field intensity is the strength of an electric field at any point. It is equal to the electric force per unit charge experienced by a test charge placed at that point.
Question 9
1 / -0

A charge $$q$$ is placed at the centre of the open end of cylindrical vessel. The flux of electric field through the surface of the vessel is

A
$$0$$

B
$$\dfrac{q}{\epsilon_0}$$

C
$$\dfrac{q}{2\epsilon_0}$$

D
$$\dfrac{2q}{\epsilon_0}$$

Solution

Given that, A charge q is placed at the center of open end Q a cylindrical vessel,we have to find the flux through the surface of the vessel.
so, when charge Q is placed at the center of open end of a cylindrical vessel then only half of the charge will contribute to the flux, because half will lie inside the surface and half will lie outside the surface.
so, flux through the surface of vessel is $$\cfrac { q }{ { 2\varepsilon }_{ 0 } } $$
Hence the answer is option (c).
Question 10
1 / -0

Two point charges each of charge $$+q$$ are fixed at $$(+a, 0)$$ and $$(-a, 0)$$. Another positive point charge $$q$$ placed at the origin is free to move along x-axis. The charge $$q$$ at the origin in equilibrium will have:

A
maximum force and minimum potential energy

B
minimum force and maximum potential energy

C
maximum force and maximum potential energy

D
minimum force and minimum potential energy

Solution

Force on q
=$$\dfrac { K{ q }^{ 2 } }{ { a }^{ 2 } } \widehat { i } -\dfrac { Kq^{ 2 } }{ { a }^{ 2 } } \widehat { i } \quad =\quad 0\\$$
potential energy
$$=\quad \dfrac { K{ q }^{ 2 } }{ { a } } \widehat { i } +\dfrac { Kq^{ 2 } }{ { a } } \widehat { i } \quad \\ =\quad \dfrac { 2K{ q }^{ 2 } }{ { a } }$$
So at the origin the force is minimum and potential energy is maximum . so the correct answe is B.

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Electric Charges and Fields Test - 45

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Electric Charges and Fields Test - 45

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Question 1

$$1$$

$$0.75$$

$$0.5$$

$$0.25$$

Solution

Question 2

$$\dfrac{1}{4 \pi \varepsilon_0} \dfrac{q}{l^2}$$

$$\dfrac{1}{4 \pi \varepsilon_0} \dfrac{q^2}{4l^2}$$

$$\dfrac{1}{4 \pi \varepsilon_0} \dfrac{q^2}{l^2}$$

$$\dfrac{1}{(4 \pi \varepsilon_0)} \dfrac{q}{l}$$

Solution

Question 3

plane, two

line, three

plane, three

line, two

Solution

Question 4

$$\dfrac{\pi}{4}$$

$$\dfrac{\pi}{3}$$

$$\dfrac{\pi}{2}$$

$${\pi}$$

Solution

Question 5

$$q/4 \pi \varepsilon _0L$$

Zero

$$q/2 \pi \varepsilon _0L$$

$$q/3 \pi \varepsilon _0L$$

Solution

Question 6

$$-0.125Nm^2/C$$

$$-0.25Nm^2/C$$

$$0.25Nm^2/C$$

$$0.125Nm^2/C$$

Solution

Question 7

$$4\sqrt {kg}\ metre$$

$$\dfrac {q}{4}\sqrt {k}\ metre$$

$$4 kq\ meter$$

$$\dfrac {4k}{q}\ metre$$

Solution

Question 8

Electric force

Electric field intensity

Electric potential

Electric power

Solution

Question 9

$$0$$

$$\dfrac{q}{\epsilon_0}$$

$$\dfrac{q}{2\epsilon_0}$$

$$\dfrac{2q}{\epsilon_0}$$

Solution

Question 10

maximum force and minimum potential energy

minimum force and maximum potential energy

maximum force and maximum potential energy

minimum force and minimum potential energy

Solution

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