Electric Charges and Fields Test

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Electric Charges and Fields Test - 68

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Weekly Quiz Competition

Question 1
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Two point charges $$q_{1}$$ and $$q_{2}$$ are $$3\ m$$ apart and their combined charge is $$20\mu C$$. If one repels the other with a force of $$0.075\ N$$, what are two charges?

A
$$10\mu C, 10\mu C$$

B
$$19\mu C, 1\mu C$$

C
$$15\mu C, 5\mu C$$

D
$$16\mu C, 4\mu C$$

Solution

$$q_1+q_2=20\mu C$$

$$\implies q_1=20-q_2$$

$$F=\dfrac{q_1q_2}{4\pi\epsilon_o R^2}$$

$$\implies 0.075=9\times 10^9\times \dfrac{q_1q_2}{3^2}$$

$$\implies q_1q_2=0.075\times 10^{-9}=75\times 10^{-12}C^2$$

$$\implies q_1q_2=75\mu C^2$$

$$\implies q_2(20-q_2)=75$$

$$\implies q_2^2-20q_2+75=0$$

$$\implies (q_2-15)(q_2-5)=0$$

$$\implies q_2=15\mu C$$

$$q_1=20-q_2=5\mu C$$

Answer-(C)
Question 2
1 / -0

The SI unit of solid angle is

A
radians

B
steradians

C
degrees

D
All of the above

Solution

The S.I unit of solid angle is steradian, or square radian (symbol: $$sr$$).
It is used in three-dimensional geometry, and is analogous to the radian, which qualifies planar angles.
Question 3
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Consider an area element $$dS$$ at a distance $$r$$ from a point P. Let $$\hat r$$ be the unit vector along the outward normal to $$dS$$. If $$\alpha$$ is the angle between $$\hat r$$ and $$dS$$, the element of the solid angle subtended by the area element at P is defined as

A
$$d\Omega = \dfrac{dS\ \sin \alpha}{r^2}$$

B
$$d\Omega = \dfrac{dS\ \tan \alpha}{r^2}$$

C
$$d\Omega = \dfrac{dS\ \cos \alpha}{r^2}$$

D
$$d\Omega = \dfrac{dS\ \alpha}{r^2}$$

Solution

Solid angle is defined as the component of area normal to radius per unit $$r^2$$.

$$d\Omega=\dfrac{dS\cos\alpha}{r^2}$$

Answer-(C)
Question 4
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A charge $$Q$$ is divided into $$2$$ parts and they are placed at a fixed distance. The force between the charges is always maximum when the charges are

A
$$\dfrac{Q}{3}, \dfrac{2Q}{3}$$

B
$$\dfrac{Q}{2}, \dfrac{Q}{2}$$

C
$$\dfrac{Q}{4}, \dfrac{3Q}{4}$$

D
$$\dfrac{Q}{5}, \dfrac{4Q}{5}$$

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, $$\dfrac{dF}{dQ_1}=0$$

$$\implies Q-2Q_1=0$$

$$\implies Q_1=\dfrac{Q}{2}$$

$$Q_2=Q-Q_1=\dfrac{Q}{2}$$

Answer-(B)
Question 5
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Two charges $$q_{1}$$ and $$q_{2}$$ are kept on the x-axis and the variation of electric field strength at different points on the x-axis is described in the adjacent figure graphically. Choose correct statement nature are magnitude of $$q_{1}$$ and $$q_{2}$$.

A
$$q_{1} + ve, q_{2} + ve, |q_{1}| > |q_{2}|$$

B
$$q_{1} + ve, q_{2} - ve, |q_{1}| > |q_{2}|$$

C
$$q_{1} + ve, q_{2} - ve, |q_{1}| < |q_{2}|$$

D
$$q_{1} - ve, q_{2} + ve, |q_{1}| < |q_{2}|$$

Solution
Question 6
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Two charges of $$-2Q$$ and $$Q$$ are located at points $$(a,0)$$ and $$(4a,0)$$ respectively. What is the electric flux through a sphere of radius $$3a$$ centred at the origin?

A
$$\dfrac{Q}{ \epsilon_0}$$

B
$$\dfrac{-Q}{ \epsilon_0}$$

C
$$\dfrac{-2Q}{ \epsilon_0}$$

D
$$\dfrac{3Q}{ \epsilon_0}$$

Solution

We know that flux$$=\phi=\dfrac{q}{\epsilon_o}$$ where $$q$$ is charge inside the closed surface area and flux due to charge outside is zero.

Here, charge inside$$=q=-2Q$$

Hence, flux$$=\phi=\dfrac{-2Q}{\epsilon_o}$$

Answer-(C)
Question 7
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Point charges $$q_1$$ and $$q_2$$ are placed in air at a distance 'r'. The ratio of the force on charge $$q_1$$ by charge $$q_2$$ and force on charge $$q_2$$ by charge $$q_1$$ is?

A
$$\displaystyle\frac{q_1}{q_2}$$

B
$$\displaystyle\frac{q_2}{q_1}$$

C
$$1$$

D
$$\left(\displaystyle\frac{q_1}{q_2}\right)^2$$

Solution

Force on charge $$q_1$$ by $$q_2$$ is given by $$F_{12} = \dfrac{1}{4\pi\epsilon_o }\dfrac{q_1q_2}{r^2}$$
Force on charge $$q_2$$ by $$q_1$$ is given by $$F_{21} = \dfrac{1}{4\pi\epsilon_o }\dfrac{q_2q_1}{r^2}$$
Thus ratio of force $$\dfrac{F_{12}}{F_{21}} = 1$$
Question 8
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An electric charge $$q$$is placed at the centre of a cube of side $$a$$ The electric flux through one of its faces is

A
$$\dfrac{q}{6\epsilon_0}$$

B
$$\dfrac{q}{6\epsilon_0a^2}$$

C
$$\dfrac{q}{6\pi \epsilon_0a^2}$$

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

Solution

Flux through the cube $$=\phi=\dfrac{q}{\epsilon_o}$$

Since, charge is placed symmetrically and is equivalent to all 6 faces , hence flux through all 6 faces will be equal and the sum of flux through all six face equals the total flux.

Hence, flux through each face $$=\dfrac{q}{6\epsilon_o}$$

Answer-(A)
Question 9
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A charge $$Q$$ is located at the centre of a sphere of radius $$R$$. Calculate the flux going out through the surface of the sphere.

A
$$\dfrac{Q}{4\pi \epsilon_0 R^2}$$

B
$$\dfrac{Q}{4\pi \epsilon_0 R}$$

C
$$\dfrac{Q}{4\pi R^2}$$

D
$$\dfrac{Q}{\epsilon_0}$$

Solution

Flux going out of a closed surface is given by charge inside closed surface divided by permittivity of free space.

$$\phi=\dfrac{Q}{\epsilon_o}$$

Answer-(D)
Question 10
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Directions For Questions

A uniformly charged conducting sphere of $$2.4m$$ diameter has a surface charge density of $$80 \mu C/m^2$$.

...view full instructions

Find the charge on the sphere

A
$$1.45\mu C$$

B
$$1.45mC$$

C
$$1.45C$$

D
$$1.45kC$$

Solution

Charge,$$q$$=surface charge density $$(\sigma) \times $$ area $$(A)$$

$$\implies q=\sigma\times 4\pi r^2$$

$$\implies q=\sigma\times 4\pi \left(\dfrac{d}{2}\right)^2$$ where $$d=diameter$$

$$\implies q=80\times 10^{-6}\times 4\pi\times \left(\dfrac{2.4}{2}\right)^2$$

$$\implies q=1.45\times 10^{-3}C=1.45mC$$

Answer-(B)

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Electric Charges and Fields Test - 68

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Electric Charges and Fields Test - 68

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Question 1

$$10\mu C, 10\mu C$$

$$19\mu C, 1\mu C$$

$$15\mu C, 5\mu C$$

$$16\mu C, 4\mu C$$

Solution

Question 2

radians

steradians

degrees

All of the above

Solution

Question 3

$$d\Omega = \dfrac{dS\ \sin \alpha}{r^2}$$

$$d\Omega = \dfrac{dS\ \tan \alpha}{r^2}$$

$$d\Omega = \dfrac{dS\ \cos \alpha}{r^2}$$

$$d\Omega = \dfrac{dS\ \alpha}{r^2}$$

Solution

Question 4

$$\dfrac{Q}{3}, \dfrac{2Q}{3}$$

$$\dfrac{Q}{2}, \dfrac{Q}{2}$$

$$\dfrac{Q}{4}, \dfrac{3Q}{4}$$

$$\dfrac{Q}{5}, \dfrac{4Q}{5}$$

Solution

Question 5

$$q_{1} + ve, q_{2} + ve, |q_{1}| > |q_{2}|$$

$$q_{1} + ve, q_{2} - ve, |q_{1}| > |q_{2}|$$

$$q_{1} + ve, q_{2} - ve, |q_{1}| < |q_{2}|$$

$$q_{1} - ve, q_{2} + ve, |q_{1}| < |q_{2}|$$

Solution

Question 6

$$\dfrac{Q}{ \epsilon_0}$$

$$\dfrac{-Q}{ \epsilon_0}$$

$$\dfrac{-2Q}{ \epsilon_0}$$

$$\dfrac{3Q}{ \epsilon_0}$$

Solution

Question 7

$$\displaystyle\frac{q_1}{q_2}$$

$$\displaystyle\frac{q_2}{q_1}$$

$$1$$

$$\left(\displaystyle\frac{q_1}{q_2}\right)^2$$

Solution

Question 8

$$\dfrac{q}{6\epsilon_0}$$

$$\dfrac{q}{6\epsilon_0a^2}$$

$$\dfrac{q}{6\pi \epsilon_0a^2}$$

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

Solution

Question 9

$$\dfrac{Q}{4\pi \epsilon_0 R^2}$$

$$\dfrac{Q}{4\pi \epsilon_0 R}$$

$$\dfrac{Q}{4\pi R^2}$$

$$\dfrac{Q}{\epsilon_0}$$

Solution

Question 10

$$1.45\mu C$$

$$1.45mC$$

$$1.45C$$

$$1.45kC$$

Solution

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