Question

Two point charges of 2esu and 3esu are placed
10 cm apart in vacuum. The electrostatic force
between them is
(1) 6 x 10-7 N
(3) 5.4 x 10-7 N
(2) 6 x 10-2 N
(4) 5 x 10-2 N​

Answer 1

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2) 6×10-2N

Answer 2

Answer:

Two-point charges of 2esu and 3esu are placed  10 cm apart in a vacuum. The electrostatic force  between them is  6 x 10-7 N

Explanation:

• The electrostatic force is the interactive force between particles at rest due to their electric charges

• When two charges are $q_{1}$ and $q_{2}$ are separated by a distance $r$ the force between the charges is given by Coulomb's law is directly proportional to the product of their charges and inversely proportional to the square of the distance between the charges

                          $F=K\frac{q_{1} q_{2} }{r^{2} }$

From the question, we have

charge of the first particle $q_{1} =2esu=2*3.33*10^{-10} C$

charge of the second particle $q_{2} =3esu=3*3.33*10^{-10}C$

(use this unit conversion $1esu=3.33*10^{-10}C$)

the distance between the charges $r=10cm=0.10m$

the value of $K=9*10^{9}$

substitute these values to get force between them

               $F=\frac{9*10^{9}*6.66*10^{-10}*9.99*10^{-10}}{0.10^{2} }$

                   $=5.98*10^{-7} N$

                   ≈$6*10^{-7}N$