Electric Charges and Fields Test

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Electric Charges and Fields Test - 24

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Question 1
1 / -0

Two uncharged bodies on rubbing, get charged due to:

A
conduction

B
friction

C
induction

D
none of these

Solution

Answer is B.

When two bodies are rubbed together, there is transfer of electrons because of friction.
Hence, two uncharged bodies on rubbing, gets equal and opposite charges.
Question 2
1 / -0

What do you understand by the static or frictional electricity ?

A
Discarging of two bodies on rubbing them together is called static or frictional electricity.

B
Charging of two bodies on rubbing them together is called static or frictional electricity.

C
Electricity which is static is called static or frictional electricity.

D
Electricity on high friction bodies is called static or frictional electricity.

Solution

Static electricity is produced when a suitable combination of substances (e.g., glass rod and silk, ebonite rod and cat's skin, sealing wax and wool) are rubbed together and get electrified due to friction.
Question 3
1 / -0

5 charges each of magnitude $$10^{-5}$$ C and mass 1 kg are place (fixed) symmetrically about a movable central charge of magnitude $$5 \times 10^{-5} C$$ and mass 0.5 kg as shown. The charge at $$P_1$$ is removed. The acceleration of the central charge is
[Given $$OP_1 = OP_2 = OP_3 = OP_4 = OP_5=1m; \displaystyle \frac{1}{4 \pi \varepsilon _0 } = 9 \times 10^9$$ in SI Units]

A
$$9 ms^{-2}$$ upwards

B
$$4.5 ms^{-2}$$ upwards

C
$$9 ms^{-2}$$ downwards

D
$$4.5 ms^{-2}$$ downwards

Solution

After removing charge from P, net force on central charge will be
$$\displaystyle F = \frac{Kq_1 q_2}{r^2} = \frac{9 \times 10^9 \times 10^{-5} \times 5 \times 10^{-5}}{12}$$
$$F = 4.5 N$$
$$\because m = 0.5 kg$$
so, acceleration,
$$\displaystyle a = \frac{F}{M} = \frac{4.5}{0.5} = 9 m/s^2 $$ upwards
Question 4
1 / -0

An electron of mass m. initially at rest takes time $$t_1$$ to move a distances in a uniform electric field in the same field environment, a proton of mass mp initially at rest takes time $$t_2$$ to move the same distance (in the opposite direction). Ignoring gravity, the ratio $$t_2/t_1$$ is

A
1

B
$$\left (\frac {m_e}{m_p}\right )^{1/2}$$

C
$$\frac {m_p}{m_e}$$

D
$$\left (\frac {m_p}{m_e}\right )^{1/2}$$

Solution

For electron, $$t_1=\sqrt {\frac {2S}{a_e}}$$
For protion, $$t_2=\sqrt {\frac {2S}{a_p}}$$
or $$\frac {t_2}{t_1}=\sqrt {\frac {a_e}{a_p}}=\sqrt {\frac {eE}{m_e}\times \frac {m_p}{eE}}$$
$$=\sqrt {\frac {m_p}{m_e}}$$
Question 5
1 / -0

Two charges are placed at certain distance apart. A metallic sheet is placed between them. What will happen to the force between the charges?

A
increases

B
decreases

C
remains the same

D
may increase or decrease depending upon the shape of the metallic sheet

Solution

Force will decreases because when medium is inserted between then it will reduces the forces by its permitivity.
the force becomes

$$F{ e }_{ m }=\quad \dfrac { F{ e } }{ { \varepsilon }_{ r } } $$

Option B is correct.
Question 6
1 / -0

An inflated balloon was pressed against a wall after it has been rubbed with a piece of synthetic cloth. It was found that the balloon sticks to the wall. What force might be responsible for the attraction between the balloon and the wall?

A
Gravitational

B
Magnetic

C
Electrostatic

D
adhesive

Solution

Electrostatic force is the phenomenon that results from slow-moving or stationary electrical charges. Specifically, electrostatic force is the physical reaction that holds together the electromagnetic field created by subatomic particles, such as electrons and protons. In order for electrostatic forces to remain cohesive, these particles need to independently maintain both positive and negative charges and react to each other accordingly.
When a balloon is rubbed with a cloth, the electrons from the atoms of the cloth enter the atoms of the balloon and the number of electrons in the atoms of balloon becomes more than the number of protons. Because of that, there is a negative charge in the balloon, and in the cloth the number of electrons in its atoms become less than the number of protons, so it's positively charged.
Therefore, due to this electrostatic force, the balloon sticks to the wall.
Question 7
1 / -0

When we remove polyester or woollen cloth in dark, we can see sparkand hear a cracking sound. These are due to:

A
static electricity

B
current electricity

C
reflection of light

D
refraction of light

Solution

The polyester or woolen cloth get rubbed with inner shirt and due to friction among these static charge gets accumulated on polyester or woolen cloth. This static charges get discharges while removing polyester or woolen cloth causing sparks and crackling sound.
Question 8
1 / -0

If two bodies A and B (A bigger in size than B) are rubbed together, then :

A
A and B get equal and opposite charges

B
A and B get equal and similar charges

C
A gets more charge than B, but of opposite kind

D
A gets less charge than B, but of same kind

Solution

Answer is A.

Static electricity is created when two objects are rubbed together, causing an object to give up or gain electrons. The electric charges that are created do not depend on the size of the objects. Therefore, when two bodies are rubbed together, both the bodies transfer electrons to and fro between the bodies and they obtain equal and opposite charges. The body that gives away electrons, gains positive charge and the body that obtains electrons gains negative charge.
Question 9
1 / -0

The adjoining figure shows a negatively charged electroscope. If a negatively charged rod is brought close to, but not touching, the knob, the two leaves will :

A
move closer together

B
move farther apart

C
not move at all

D
None of these

Solution

Answer is B.

In this case, the given electroscope is negatively charged. If a negatively charged rod is brought close to, but not touching, the knob, the two leaves will move far apart as the like charges repel each other.
Question 10
1 / -0

12 positive charges of magnitude q are placed on a circle of radius R in a manner that they are equally spaced. A charge +Q is placed at the centre. If one of the charges q is removed, then the force on Q is

A
zero

B
$$\displaystyle \frac{qQ}{4 \pi \varepsilon_0 R^2}$$ away from the position of the removed charge.

C
$$\displaystyle \frac{11 qQ}{4 \pi \varepsilon_0 R^2}$$ away from the position of the removed charge.

D
$$\displaystyle \frac{qQ}{4 \pi \varepsilon_0 R^2}$$ towards the position of the removed charge.

Solution

$$\displaystyle \frac{KQq}{R^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{R^2}$$
Since initially net force on Q was zeroby symmetry
So, $$\bar{F_1} + \bar{F}_{Remaining 11} = 0$$
$$\bar{F}_{Remaining 11} = - \bar{F_1}$$
So, $$\displaystyle \frac{1}{4 \pi \varepsilon_0} . \frac{Qq}{r^2}$$ towards the position of the removed charge.

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Electric Charges and Fields Test - 24

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Electric Charges and Fields Test - 24

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Question 1

conduction

friction

induction

none of these

Solution

Question 2

Discarging of two bodies on rubbing them together is called static or frictional electricity.

Charging of two bodies on rubbing them together is called static or frictional electricity.

Electricity which is static is called static or frictional electricity.

Electricity on high friction bodies is called static or frictional electricity.

Solution

Question 3

$$9 ms^{-2}$$ upwards

$$4.5 ms^{-2}$$ upwards

$$9 ms^{-2}$$ downwards

$$4.5 ms^{-2}$$ downwards

Solution

Question 4

1

$$\left (\frac {m_e}{m_p}\right )^{1/2}$$

$$\frac {m_p}{m_e}$$

$$\left (\frac {m_p}{m_e}\right )^{1/2}$$

Solution

Question 5

increases

decreases

remains the same

may increase or decrease depending upon the shape of the metallic sheet

Solution

Question 6

Gravitational

Magnetic

Electrostatic

adhesive

Solution

Question 7

static electricity

current electricity

reflection of light

refraction of light

Solution

Question 8

A and B get equal and opposite charges

A and B get equal and similar charges

A gets more charge than B, but of opposite kind

A gets less charge than B, but of same kind

Solution

Question 9

move closer together

move farther apart

not move at all

None of these

Solution

Question 10

zero

$$\displaystyle \frac{qQ}{4 \pi \varepsilon_0 R^2}$$ away from the position of the removed charge.

$$\displaystyle \frac{11 qQ}{4 \pi \varepsilon_0 R^2}$$ away from the position of the removed charge.

$$\displaystyle \frac{qQ}{4 \pi \varepsilon_0 R^2}$$ towards the position of the removed charge.

Solution

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