Electric Charges and Fields Test

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Electric Charges and Fields Test - 57

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Question 1
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Force between two charges $$q_1$$ and $$q_2$$ separated by a distance r is proportional to $$q _ { 1 } q _ { 2 } / r ^ { 2 }$$ . Proportionality constant is

A
$$\dfrac { \varepsilon _ { 0 } } { 4 \pi }$$

B
$$4 \pi \varepsilon _ { 0 }$$

C
$$\dfrac { 1 } { 4 \pi \varepsilon _ { 0 } }$$

D
$$1$$

Solution

Given that,

First charge $$={{q}_{1}}$$

Second charge $$={{q}_{2}}$$

Distance $$=r$$

We know that,

According to Coulomb’s law

The force is

$$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\times \dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$$

So, the proportionality constant is

$$=\dfrac{1}{4\pi {{\varepsilon }_{0}}}$$

Hence, this is the required solution
Question 2
1 / -0

Eight point charges (can be assumed as small spheres uniformly charged and their centres at the corner of the cube) having values q each are fixed at vertices of a cube. The electric flux through square surface ABCD of the cube is

A
$$\frac {q}{24 \in_0}$$

B
$$\frac {q}{12 \in_0}$$

C
$$\frac {q}{6 \in_0}$$

D
$$\frac {q}{8 \in_0}$$

Solution

Using Gauss law we said that,
$$\phi =\dfrac { q }{ { \epsilon }_{ 0 } } $$
Since, all the c/s are in the $$8$$ corners, Then,
$$\phi =\dfrac { 1 }{ 8 } \dfrac { q }{ { \epsilon }_{ 0 } } $$
Question 3
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A charge +Q is located in space at point $$(x=1 m,y=10m, z=5m)$$. What is the total electric flux that passes through the $$y-z$$ plane?

A
$$\cfrac Q {\varepsilon_0}$$

B
$$\cfrac Q {3\varepsilon_0}$$

C
$$\cfrac Q {6\varepsilon_0}$$

D
$$\cfrac Q {2\varepsilon_0}$$

Solution

A charge $$+Q$$
Point ($$x=1m$$, $$y=10m$$, $$z=5m$$)
Using the Gaussian theorem,
total electric flux $$=\phi =\dfrac { Q }{ { \epsilon }_{ 0 } } $$
Question 4
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A light beam travelling in the x-direction is described by the electric field $${ E }_{ y }=(300V{ m }^{ -1 })sin\quad \omega (t-x/c).$$ An electron is constrained to move along the y-direction with a speed of $$2.0\times {10}^7 { m }^{ -1 }$$ Find the maximum electric force and the maximum magnetic force on the electron

A
$$4.8\times {10}^{-17} N, zero$$

B
$$4.2\times {10}^{-18} N, 1.8\times 10^{-8}N$$

C
$$4.8\times {10}^{-17} N, 3.2\times 10^{-18}N$$

D
Zero, Zero

Solution

A light beam travelling
$${ E }_{ y }=\left( 300V/m \right) sinw\left( t-x/c \right) $$.
$$V=2.0\times { 10 }^{ 7 }{ m }^{ -1 }$$
$${ B }_{ 0 }=\dfrac { { F }_{ 0 } }{ C } =\dfrac { 300 }{ 2\times { 10 }^{ 7 } } =150\times { 10 }^{ -7 }T$$
$${ F }_{ m }={ B }_{ 0 }qV$$
$$=9.1\times { 10 }^{ -31 }\times 150\times { 10 }^{ -7 }\times 2\times { 10 }^{ 7 }$$
$$=2.8\times { 10 }^{ -28 }\times 2$$
$$=4.8\times { 10 }^{ -17 }N$$
magnetic force $$=0$$
because they are perpendicular to each other.
Question 5
1 / -0

Choose the correct answer (Only one option is correct) from the following.
Statement-1: A deuteron and $$\alpha - particle$$ are placed in an electric field. If $${F_1}\;{\text{and}}\;{F_2}$$ be the forces acting on them and $${a_1}\;{\text{and}}\;{a_2}$$ be their accelerations respectively, then $${a_1} = {a_2}.$$.

Statement-2: Forces will be same in the electric field.

A
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

B
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

C
Statement-1 is true, Statement-2 is false

D
Statement-1 is false , Statement-2 is true

Solution
Question 6
1 / -0

Flux through the rectangular piece of area 10 cm by 20 cm when placed ( as shown in an uniform electric field of 200 N/C, ) is

A
4.0 $$Nm^{2}/C$$

B
3.5 $$Nm^{2}/C$$

C
2.5 $$Nm^{2}/C$$

D
zero

Solution

$$E=3\times { 10 }^{ 3 }N/C$$
and square is parallel to $$y-z$$ plane, So, flux will be simply one multiplication of area of electric field which is as follows.
$$=3\times \left( { 10 }^{ 3 } \right) \times 20\times \left( { 10 }^{ -9 } \right) $$
$$=3Vm$$
Question 7
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Two parallel plates have charges $$Q_1$$ and $$Q_2$$ on them and the capacitance is c. The potential difference between them is

A
$$\dfrac{Q_1+Q_2}{2C}$$

B
$$\dfrac{Q_1-Q_2}{2C}$$

C
$$\dfrac{Q_1+Q_2}{C}$$

D
$$\dfrac{Q_1-Q_2}{C}$$

Solution

$$x+y=Q_1$$
$$-x+y=Q_2$$
So, $$x+x=Q_1-Q_2\implies x=\cfrac{Q_1-Q_2}{2}$$
Potential difference$$=\cfrac{x}{c}\\=\cfrac{Q_1-Q_2}{2C}$$
Question 8
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As shown in figure a dust particle with mass $$m=5.0\times 10^{-9}$$ kg and charge $$q_{0}=2.0 nC$$ starts from rest at point a and moves in a straight line to point b. What is its speed v at point b?

A
26 $$ms^{-1}$$

B
34 $$ms^{-1}$$

C
46 $$ms^{-1}$$

D
14 $$ms^{-1}$$
Question 9
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Four charges $$+Q,-Q,+Q$$ and $$-Q$$ are situated at the corners of a square; in a sequence then at the centre of the square:

A
$$E=0,V=0$$

B
$$E=0,V\neq 0$$

C
$$E\neq 0,V=0$$

D
$$E\neq 0,V\neq 0$$

Solution

Four charges $$+Q$$, $$-Q$$, $$+Q$$, $$-Q$$.
In this context, $$E\neq 0$$ but $$V=0$$ because. $$E$$ is not cancel out to each other but $$V$$ is cancel out each other.
Question 10
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There are two charges $$+2\mu\ C\ and -3\mu\ C$$. The ratio of forces acting on them will be

A
$$2:3$$

B
$$1:1$$

C
$$3:2$$

D
$$4:9$$

Solution