Question

A small glass ball rubbed with silk gains a charge of +2.0 μC. The glass ball is placed 12 cm from a small charged rubber ball that carries a charge of −3.5 μC. a. What is the magnitude of the electric force between the two balls? b. Is this force attractive or repulsive? c. How many electrons has the glass ball lost in the rubbing process?

Answer 1

Answer:

(A) The magnitude of force is 4.375 N

(B) Force is attractive

(C) The electron lost in rubbing process is $1.25 \times 10^{19}$

Explanation:

Given:

Slik gain charge $q_{1} = +2\mu C$

Glass ball charge $q_{2} = -3.5 \mu C$

Separation between two charge $r = 0.12$ m

(A)

From the coulomb's law,

  $F = \frac{k q_{1} q_{2} }{r^{2} }$

Where $k= 9 \times 10^{9}$

$F = 9 \times 10^{9} (\frac{2 \times -3.5 \times 10^{-12} }{(0.12)^{2} } )$

$F = -4.375$ N

But magnitude of force is given by,

$F = 4.375$ N

(B)

Here, force is attractive because positive charge attract negative charge

(C)

The electron lost in rubbing process is

$2 = n (1.6 \times 10^{-19} )$

$n = 1.25 \times 10^{19}$

Therefore, the electron lost in rubbing process is $1.25 \times 10^{19}$ and magnitude of force is 4.375 N and force is attractive.

Answer 2

a) The magnitude of the electric force between the two balls is 4.375 N.

b) Force is attractive.

c) The electrons lost in rubbing process is 1.25 x 10¹³.

What is electrostatic force?

The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

F = kq₁q₂/r²

where k = 9 x 10⁹ N.m²/C²

Given is the silk gain charge q₁= 2 x 10⁻⁶ C and glass ball charge q₂= 3.5x 10⁻⁶ C, distance between charges r =12 cm =0.12m

To find the magnitude of force, only the magnitude of charges are taken.

Substituting the values into the expression ,we get the electrostatic force

F =  (9 x 10⁹ x 2 x 10⁻⁶ x 3.5x 10⁻⁶)/ (0.12)²

F = 4.375 N

Thus, the magnitude of the electric force between the two balls is 4.375 N.

b) The force is attractive because the positive charge attracts the  negative charge.

c) The electron lost in rubbing process is

q₁ =ne

where charge on electron, e = 1.6 x 10⁻¹⁹C and n is the number of electrons.

Put the values, we get

2 x  10⁻⁶ = n x  1.6 x 10⁻¹⁹

n = 1.25 x 10¹³

Therefore, the electron lost in rubbing process is 1.25 x 10¹³.

Learn more about electrostatic force.

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