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Grade: 11
        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?
5 years ago

Answers : (4)

Aditi Chauhan
askIITians Faculty
396 Points
							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

						
5 years ago
Apoorva Arora
IIT Roorkee
askIITians Faculty
181 Points
							
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
5 years ago
Anil sharma
17 Points
							The charge q1 & q2 are placed at corner of square find q2 such that the resultant force on q1 is zero
						
3 years ago
Diksha
11 Points
							
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
2 years ago
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