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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?
5 years ago

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
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 valuesF= $2.304\times 10^{-24} N$Thanks and RegardsApoorva AroraIIT RoorkeeaskIITians 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/root20=kqQ/a^2+kqQ/2root2a^2qQ+q^2/2root2=0.        q= -2root2Q

2 years ago
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• 57 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions