Electric Charges and Fields Test

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Electric Charges and Fields Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

A and B are two points on the axis and the perpendicular bisector respectively of an electric dipole. A and B are far away from the dipole and at equal distances from its centre. The fields at A and B i.e. $$\overrightarrow { { E }_{ A } } \quad and\overrightarrow { { E }_{ B } } $$ are respectively such that

A
$$\overrightarrow { { E }_{ A } } =\overrightarrow { { E }_{ B } } $$

B
$$\overrightarrow { { E }_{ A } } =2\overrightarrow { { E }_{ B } } $$

C
$$\overrightarrow { { E }_{ A } } =-2\overrightarrow { { E }_{ B } } $$

D
$$\overrightarrow { { E }_{ A } } =\frac { 1 }{ 2 } \overrightarrow { { E }_{ B } } $$

Solution

It will be cleared Heat electric field dual to an electric dipole at axial point is $$2Kp/{ r }^{ 3 }$$ while at a point on perpendicularity bisector is $$Kp/{ r }^{ 3 }$$. So, if electric field at point $$A$$ is $$E$$ then electric field at point $$B$$ will be $$E/2$$.
Now, $${ E }_{ A }=-2{ E }_{ B }$$
Question 2
1 / -0

An electric charge produces an electric intensity of $$500 N/C$$ at a point in air. If the air is replaced by a medium of dielectric constant $$2.5$$, then the intensity of the electric field due to the same charge at the same point will be:

A
$$100N/C$$

B
$$150N/C$$

C
$$200N/C$$

D
$$300N/C$$

Solution

Electric intensity $$=500N/C$$
$$K=2.5$$
Intensity of electric field due to the same charge at same point will be
$$\phi =\dfrac { q }{ { \epsilon }_{ 0 } } $$ (using Gauss law)
or, $$\phi =\dfrac { 500 }{ 2.5 } =\dfrac { 500 }{ 25 } \times 10=150N/C$$
Question 3
1 / -0

A particle A has charge +q and a particle B has charge +4q. each having the same m . allowed to fall from rest through the same electric potential difference. the ratio of the speed of A to that of B will e

A
2 :1

B
1 :2

C
4 :1

D
1 :4

Solution

Particle A Particle B
Charge =q Charge =4q
mass=m mass=m
potential=$$V_A$$ potential=$$V_B$$
Now, kinetic energy, $$K=$$charge $$\times $$potential
$$=qv$$
and kinetic energy $$ = \frac{1}{2}m{v^2}$$
$$\begin{array}{l} \therefore \frac { { { K_{ A } } } }{ { { K_{ B } } } } =\dfrac { { qv } }{ { 4qv } } =\dfrac { { \dfrac { 1 }{ 2 } mv_{ A }^{ 2 } } }{ { \dfrac { 1 }{ 2 } mv_{ B }^{ 2 } } } \\ or\, \, \, ,\dfrac { { { V_{ A } } } }{ { { V_{ B } } } } =\dfrac { 1 }{ 2 } \end{array}$$
Hence Option $$B$$ is correct .
Question 4
1 / -0

Force between two charges at rest is____

A
electrostatic force

B
elector magnetic force

C
real force

D
dynamic force

Solution

Force between two charges at rest is electrostatic force of coulomb's law.
Question 5
1 / -0

There are two $$ +2 \mu C$$ and$$ -3 \mu C $$ some distance apart. The ratio of force acting on them will be

A
2 :3

B
1 : 1

C
3 : 2

D
4 : 9

Solution
Question 6
1 / -0

Three charges $$+Q, q, + Q$$ are placed respectively, at distance, $$0, d/2$$ and $$d$$ from the origin, on the x-axis. If the net force experienced by $$+ Q$$, placed at $$x= 0$$, $$Ls$$ zero,then value of $$q$$ is :

A
$${+Q}/{2}$$

B
$${-Q}/{4}$$

C
$${+Q}/{4}$$

D
$${-Q}/{2}$$

Solution

For equilibrium,
$$\overrightarrow{F}_a + \overrightarrow{F}_b=0$$

$$\overrightarrow{F}_a=-\overrightarrow{F}_b$$

$$\dfrac{kQQ}{d^2}=-\dfrac{kQq}{(d/2)^2}$$

$$\Rightarrow q = -\dfrac{Q}{4}$$
Question 7
1 / -0

The force of attraction between two charges $$8\mu C$$ and $$0.2 \,N$$. Find the distance of separation.

A
$$412$$ meter

B
$$1.2$$ meter

C
$$120$$ meter

D
$$0.12$$ meter

Solution

$$F = \dfrac{K \,q_1q_2}{r^2}$$

$$r^2 = \dfrac{9 \times 10^9 \times 8 \times 10^{-6} \times 4 \times 10^{-6}}{r^2}$$

$$r^2 = \dfrac{9 \times 8 \times 4}{2} \times 10^{-3}$$

$$r^2 = \dfrac{9 \times 8 \times 4}{2} \times 10^{-2}$$

$$r = 12 \times 10^{-1}$$ meter $$= 1.2 \,m$$
Question 8
1 / -0

Which of the following is not a unit of charge ?

A
coulomb

B
ampere-second

C
microcoulomb

D
ampere/second

Solution

(i) The charge unit coulomb
(ii) $$q=it=$$ ampere $$\times$$ second
(iii) micro coulomb (because $$s$$ prefix only is added)
(iv) ampere/second $$\rightarrow$$ it is not unit of charge.
Question 9
1 / -0

If point charges $$Q_1=2\times 10^{-7}C$$ and $$Q_2=3\times 10^{-7}C$$ are at. $$30cm$$ separation. Find electrostatic force them

A
$$6\times 10^{-3}N$$

B
$$2\times 10^{-3}N$$

C
$$3\times 10^{-3}N$$

D
$$8\times 10^{-3}N$$

Solution
Question 10
1 / -0

Two charged particles having charges $$+ 25 \mathrm { mC }$$ and $$+ 50 \mathrm { mC }$$ are separated by a distance of 8 cm. The ratio of force on them is:

A
$$1 : 1$$

B
$$1 : 2$$

C
$$4 : 1$$

D
$$2 : 1$$

Solution

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Particle A	Particle B
Charge =q	Charge =4q
mass=m	mass=m
potential=$$V_A$$	potential=$$V_B$$

Electric Charges and Fields Test - 59

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Electric Charges and Fields Test - 59

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Question 1

$$\overrightarrow { { E }_{ A } } =\overrightarrow { { E }_{ B } } $$

$$\overrightarrow { { E }_{ A } } =2\overrightarrow { { E }_{ B } } $$

$$\overrightarrow { { E }_{ A } } =-2\overrightarrow { { E }_{ B } } $$

$$\overrightarrow { { E }_{ A } } =\frac { 1 }{ 2 } \overrightarrow { { E }_{ B } } $$

Solution

Question 2

$$100N/C$$

$$150N/C$$

$$200N/C$$

$$300N/C$$

Solution

Question 3

2 :1

1 :2

4 :1

1 :4

Solution

Question 4

electrostatic force

elector magnetic force

real force

dynamic force

Solution

Question 5

2 :3

1 : 1

3 : 2

4 : 9

Solution

Question 6

$${+Q}/{2}$$

$${-Q}/{4}$$

$${+Q}/{4}$$

$${-Q}/{2}$$

Solution

Question 7

$$412$$ meter

$$1.2$$ meter

$$120$$ meter

$$0.12$$ meter

Solution

Question 8

coulomb

ampere-second

microcoulomb

ampere/second

Solution

Question 9

$$6\times 10^{-3}N$$

$$2\times 10^{-3}N$$

$$3\times 10^{-3}N$$

$$8\times 10^{-3}N$$

Solution

Question 10

$$1 : 1$$

$$1 : 2$$

$$4 : 1$$

$$2 : 1$$

Solution

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