Question

calculate the force between two charges of 1c separated by 10m in air.plz answer fast
​

Answer 1

Answer:

Force =

r

2

kq

1

q

2

=

(20×10

−2

)

2

9×10

9

×20×10

−3

×20×10

−3

=

4×10

−2

36×10

5

=9×10

7

N

Answer 2

Answer:

Two charges 1C and 1C are kept separated by a distance of 10m in the air the magnitude of the force between them​ is $9*10^{7}N$

Explanation:

• The interaction between two charges at rest is called electrostatic force

• The coulomb's law gives the magnitude of the force in the air as

                     $F=\frac{q_{1} q_{2} }{4\pi \varepsilon _0 r^{2} }$

       where $q_{1} ,q_{2}$- the charges

       $\frac{1}{4\pi \varepsilon _0}=9*10^{9}$

       $r$- separation between the charges

From the question, we have

the charges are $q_{1} =1C ,q_{2} =-1C$

the distance between the charges $r=10m$

substitute the values, we get

$/F/=\frac{9*10^{9}*1*1 }{10^{2} }$

⇒ $/F/=9*10^{7} N$