Question

How many electron should be added or removed from a neutral body of mass 10 mg so that it may remain stationary in air in an electric field of strength 100 Newton per coulomb directed upward G is equal to 10

Answer 1

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●GIVEN :-

• Mass = 10 × 10^-6 = 10^-5 gm

• E = 100N/C

● WE HAVE :-

• g =10m/sec^2

• e = 1.6 × 10^-19

●TO FIND:-

• Number of electrons.

●FORMULA USED:-

• qE = mg

• n = q/e

● SOLUTION: -

$\sf : \longrightarrow{qE = mg} \\ \\ \sf : \longrightarrow{q \: = \frac{mg}{E} } \\ \\ \sf : \longrightarrow{q = \frac{ {10}^{ - 5} \times 1 \cancel{0 }}{10 \cancel{0}} } \\ \\ \sf : \longrightarrow{q = {10}^{ - 5} \times {10}^{ - 1} } \\ \\ \sf : \longrightarrow{q = {10}^{ - 6 }C}$

Now, Number of electrons = q/e

$\sf : \longrightarrow{n = \frac{q}{e} } \\ \\ \sf : \longrightarrow{n = \frac{ {10}^{ - 6} }{1.6 \times {10}^{ - 19} } } \\ \\ \sf : \longrightarrow{n = {10}^{ - 6 + 19} \times 0.625} \\ \\ \sf : \longrightarrow \pink{n = 6.25 \times {10}^{12} electrons}$

So, 6.25 ×10^12 electrons should be removed.

Answer 2

Given:

• Mass = 10 mg = 10^(-5) kg
• Field intensity = 100 N/C
• g = 10 m/s²
• e = 1.6 × 10^(-19) C

To find:

Number of electrons needed ?

Calculation:

The gravitational force should be balanced by electrostatic field force for the body to be in equilibrium.

$E \times q = m \times g$

$\implies \: q = \dfrac{m \times g}{E}$

$\implies \: q = \dfrac{ {10}^{ - 5} \times 10}{100}$

$\implies \: q = {10}^{ - 6} \: C$

So, number of electrons needed:

$\implies \: e = \dfrac{q}{1.6 \times {10}^{ -1 9} }$

$\implies \: e = \dfrac{ {10}^{ - 6} }{1.6 \times {10}^{ -1 9} }$

$\implies \: e = 6.25 \times {10}^{12}$

So, 6.25 × 10¹² electrons are needed.