# 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?

Grade:11

## 4 Answers

Aditi Chauhan
askIITians Faculty 396 Points
9 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 IIT Roorkee
askIITians Faculty 181 Points
9 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
17 Points
6 years ago
The charge q1 & q2 are placed at corner of square find q2 such that the resultant force on q1 is zero
Diksha
11 Points
5 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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