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Question: The electrostatic attracting force on a small sphere of charge \[0.2\mu C\] due to another small sph...

The electrostatic attracting force on a small sphere of charge 0.2μC0.2\mu C due to another small sphere of charge −0.4μC - 0.4\mu C in air is 0.4N0.4N . The distance between two spheres is
A. 43.2×10−3m43.2 \times {10^{ - 3}}m
B. 42.4×10−3m42.4 \times {10^{ - 3}}m
C. 18.1×10−3m18.1 \times {10^{ - 3}}m
D. 19.2×10−6m19.2 \times {10^{ - 6}}m

Explanation

Solution

Use the Coulomb’s law to find the distance between the two charged particles. This law is applicable when the two charges are placed together in the presence of an electrostatic field.
F=q1q24π∈0r2F = \dfrac{{{q_1}{q_2}}}{{4\pi { \in _0}{r^2}}}

Complete step by step solution:
Here,q1=0.2μC=0.2×10−6C{q_1} = 0.2\mu C = 0.2 \times {10^{ - 6}}C and q2=−0.4μC=−0.4×10−6C{q_2} = - 0.4\mu C = - 0.4 \times {10^{ - 6}}C
F=0.4NF = 0.4N, We have to find r , i.e. the distance between them
As we know that the electrostatic force of attraction between two charges is given by Coulomb's law. According to Coulomb’s law the force of interaction between any two point charges is directly proportional to the product of the charges and inversely proportional to the distance between them. So,
F = \dfrac{{{q_1}{q_2}}}{{4\pi { \in _0}{r^2}}}$$$$ \Rightarrow {r^2} = \dfrac{{{q_1}{q_2}}}{{4\pi { \in _0}F}}
Where, ∈0{ \in _0} is the permittivity of free space. The above equation is valid for any sign of q1_1 and q2_2.

In case of negative charges only consider their magnitude. The nature of charges only gives the nature of force, i.e. attractive or repulsive acting between them.
⇒r2=(0.2×10−6)(0.4×10−6)(9×109)0.4\Rightarrow {r^2} = \dfrac{{\left( {0.2 \times {{10}^{ - 6}}} \right)\left( {0.4 \times {{10}^{ - 6}}} \right)\left( {9 \times {{10}^9}} \right)}}{{0.4}} As the value of 14π∈0=9×109\dfrac{1}{{4\pi { \in _0}}} = 9 \times {10^9}
⇒r2=1.8×10−3m\Rightarrow {r^2} = 1.8 \times {10^{ - 3}}m
⇒r2=18×10−4m\Rightarrow {r^2} = 18 \times {10^{ - 4}}m
⇒r=4.24×10−2m\Rightarrow r = 4.24 \times {10^{ - 2}}m
⇒r=42.4×10−3m\Rightarrow r = 42.4 \times {10^{ - 3}}m
Thus,the distance between the given particles is 42.4×10−3m42.4 \times {10^{ - 3}}m .

Hence, the correct option is (B).

Note: First convert all the units in SI system before putting in the formula. From the given charges we can see that force exerted by the two charges are attractive in nature. Coulomb’s law of electrostatic force between two charges corresponds to Newton’s law of gravitational force between two masses. This is as if forces are equal and opposite and Newton’s third law of motion is obeyed. Coulomb’s forces of interaction are parallel to the line joining the centers of two charged bodies. Hence electrostatic forces are conservative forces.