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Two charged particles are placed at a distance 1.0 cm apart. What is the minimum possible magnitude of the electric force acting on each charge?

Simran Bhatia , 10 Years ago
Grade 11
anser 4 Answers
Aditi Chauhan

Last Activity: 10 Years ago

Sol. Minimum charge of a body is the charge of an electron Wo, q = 1.6 × 10^–19 c x = 1 cm = 1 × 10^–2 cm So, F = kq base 1q base 2/r^2 = 9 * 10^9 * 1.6 * 1.6 * 10^-19/10^-2 * 10^-2 = 23.04 * 10^-38+9+2+2 = 23.04 * 10^-25 = 2.3 * 10^-24

Apoorva Arora

Last Activity: 10 Years ago

the distance is constant so force depends only on charge. so the force is minimum when charge is minimum i.e. charge of one electron.
So
F=\frac{kq_{1}q_{2}}{r^{2}}
putting all the values
F= 2.304\times 10^{-24} N
Thanks and Regards
Apoorva Arora
IIT Roorkee
askIITians Faculty

Anil sharma

Last Activity: 8 Years ago

The charge q1 & q2 are placed at corner of square find q2 such that the resultant force on q1 is zero

Diksha

Last Activity: 6 Years ago

F on q= vector of (F1+F2+F3)
0=F1+F2+F3(vector)  
=kqQ/a^2(i+j)+kq^2/root2a^2 i+j/root2
0=kqQ/a^2+kqQ/2root2a^2
qQ+q^2/2root2=0.        q= -2root2Q

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