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Grade 11Electromagnetic Induction

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?

Profile image of Simran Bhatia
12 Years agoGrade 11
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4 Answers

Profile image of Aditi Chauhan
12 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
Profile image of Apoorva Arora
12 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
Profile image of Anil sharma
9 Years ago
The charge q1 & q2 are placed at corner of square find q2 such that the resultant force on q1 is zero
Profile image of Diksha
8 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