Tag: comparing force in magnetic, electric and gravitational fields

Questions Related to comparing force in magnetic, electric and gravitational fields

Consider two conducting plates $A$ and $B$, between which the potential difference is $5 V$, plate $A$ being at a higher potential. A proton and an electron are released at plates $A$ and $B$ respectively. The two particles then move towards the opposite plates - the proton to plate $B$ and the electron to plate $A$. Which one will have a larger velocity when they reach their respective destination plates?

  1. Both will have the same velocity

  2. The electron will have the larger velocity

  3. The proton will have the larger velocity

  4. None will be able to reach the destination point


Correct Option: B
Explanation:

Between two conducting plates, there is the uniform electric field and therefore the same charge is applicable on both proton and neutron. Since Proton mass is more than that of the electron, it will have less acceleration and hence less speed achieved by it. Therefore electron will reach plate with larger velocity.

$F _g$ and $F _e$ represent gravitational and electrostatic force respectively between electrons situated at a distance $0.1\,m. F _g/F _e$ is of the order

  1. $10^{-41}$

  2. $10^{-45}$

  3. $10^{40}$

  4. $10^{-42}$


Correct Option: D
Explanation:
${ F } _{ y }$ and ${ F } _{ e }$ represent gravitational and electrostatic force respectively between electrons situated at a distance $=0.1m$
let, us take the ratio
$\dfrac { { F } _{ e } }{ { F } _{ g } } =\dfrac { \dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \dfrac { { q }^{ 2 } }{ { d }^{ 2 } }  }{ G\dfrac { { m }^{ 2 } }{ { d }^{ 2 } }  } $
$\dfrac { { F } _{ e } }{ { F } _{ g } } =\dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \times \dfrac { 1 }{ G } \times \dfrac { { q }^{ 2 } }{ { m }^{ 2 } } $
$\dfrac { { F } _{ e } }{ { F } _{ g } } =9\times { 10 }^{ 9 }\times \dfrac { 1 }{ 6.67\times { 10 }^{ -11 } } \times \dfrac { { \left( 1.6\times { 10 }^{ -19 } \right)  }^{ 2 } }{ { \left( 9.1\times { 10 }^{ -31 } \right)  }^{ 2 } } $
$\Rightarrow \dfrac { { F } _{ e } }{ { F } _{ g } } =4.17\times { 10 }^{ 42 }$
or,  $\dfrac { { F } _{ e } }{ { F } _{ g } } $ is of the order of ${ 10 }^{ -42 }$.

Four point charge of $+10^{-7}C, -10^{-7}C, -2\times 10^{-7}C$ and $+2\times 10^{-7}C$ are placed respectively at the corners $A, B, C, D$ of a $0.05\ m$ square. Find the magnitude of the resultant force on the charge at $D$.

  1. $0.2\ dyne$

  2. $0.2\ newton$

  3. $2\ dyne$

  4. $0.02\ newton$


Correct Option: A

An $\alpha$-particle is the nucleus of a helium atom. It has a mass $m=6.64 \times 10^{-27}$kg and a charge $q = + 2e =  3.2 \times 10^{-19}$ C. Compare the force of the electric repulsion between two a-particles with the force of gravitational attraction between them.

  1. $\displaystyle \dfrac{F _e}{F _g} = 3.1 \times 10^{32}$

  2. $\displaystyle \dfrac{F _e}{F _g} = 3.1 \times 10^{35}$

  3. $\displaystyle \dfrac{F _e}{F _g} = 3.1 \times 10^{38}$

  4. $\displaystyle \dfrac{F _e}{F _g} = 1 \times 10^{32}$


Correct Option: B
Explanation:

Electrostatic repulsion = $F _e=\dfrac{1}{4 \pi \epsilon } \dfrac{q^2}{r^2 }=\dfrac{1}{4 \pi \epsilon } \dfrac{4e^2}{r^2 } $

gravitational attraction = $ F _g = G\dfrac{m^2}{r^2} $

$\dfrac{F _e}{F _g} = \dfrac{1}{4 \pi \epsilon } \dfrac{4 e^2}{G m^2} = 3.1 \times 10^{35} $

Electrostatic force and gravitational force differ in which respect?

  1. Conservative force

  2. Central force

  3. Principle of superposition

  4. Dependence on the intervening medium


Correct Option: D
Explanation:

In electrostatic force, the force of medium depend on charges while the force of medium due to gravity does not depend on masses. There is no concept of induced mass.

The ratio of electric force between two electrons to the gravitational force between them is of the order

  1. 10$^{42}$

  2. 10$^{39}$

  3. 10$^{36}$

  4. 1


Correct Option: A
Explanation:

The electric force between two electrons is $F _e=\dfrac{ke^2}{r^2}$

and gravitational force between two electrons is $F _g=\dfrac{Gm _e^2}{r^2}$
Thus, $\dfrac{F _e}{F _g}=\dfrac{ke^2}{Gm _e^2}=\dfrac{(9\times 10^9)(1.6\times 10^{-19})^2}{(6.67\times 10^{-11})(9.1\times 10^{-31})^2}=4.17\times 10^{42} \sim 10^{42}$

During winter, clothes tend to stick to our body as an action of ______ force.

  1. electrostatic 

  2. muscular

  3. magnetic

  4. contact


Correct Option: A
Explanation:

Due to electrostatic force clothes tends to stick to our body. 

When an object is rubbed with some materials, it acquires charges and becomes ________ object.

  1. uncharged

  2. charged

  3. neutral

  4. heavy


Correct Option: B
Explanation:

When object is rubbed with some materials it acquires charges. This object is called as charged object and develops electrostatic force.

Gravitational force is a/an ______ force.

  1. repulsive

  2. attractive

  3. contact

  4. neutral


Correct Option: B
Explanation:

Gravitational force is an attractive force as it tends to move all objects towards itself.

Under action of which force does water flow from an open tap towards ground?

  1. Electrostatic force

  2. Magnetic force

  3. Gravitational force

  4. None of the above


Correct Option: C
Explanation:

Objects tend to move downwards to the earth because of gravity of the earth. This force, exerted on the object is known as force of gravity.