Question

Ex. 2.1.3
Two charges of 4 micro-coulomb and 16 micro-
coulomb are placed 40 cm apart in air. Calculate the
force between them.​

Answer 1

Answer:

electric field is zero between two changes means at point electric field of both charge same . let 60−rbe distance from 36 micro c charge so

k×4/r

2

=k×36/(60−r)

2

r

2

1

=

(60−r)

2

9

take sq root both side

r

1

=

(60−r)

3

60-r=3r

60=4r

r=15

zero field from lager charge is 60−15=45cm

Answer 2

Given:

Two charges $4$ μ$C$  and  $16$μ$C$

Distance between the charges $=40cm$

To find:

The force between the charges

Solution:

• The Coulomb's law in electrostatics states that the force between any two charges is directly proportional to the product obtained of their individual magnitude of charges and inversely proportional to the square of the distance between their centers.

                                    $F=K\frac{Q_{1} Q_{2} }{R^{2} }$

We have,

$Q_{1}=4$ × $10^{-6}C$

$Q_{2} =16$ × $10^{-6}C$

$R=40 cm=40$ × $10^{-2}m$

$K=9$ × $10^{9}Nm^{2}/C^{2}$

Hence,

$F=\frac{(9*10^{9} )(4*10^{-6} )(16*10^{-6} )}{(40*10^{-2} )^{2} }$

$F=\frac{576*10^{(9-6-6)} }{16*10^{-2} }$

$F=36$ ×$10^{-3+2}$

$F=3.6N$

Final answer:

Hence, the force between the two charges is 3.6 N.