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A charged particle of mass 0.003g is held stationary in space by placing it in a downward direction of electric field of 6x10^4 N/C. Then, the magnitude of charge is?

Vaibhavi Singh , 8 Years ago
Grade 12
anser 1 Answers
Ankit Kumar
m=0.003g=3×10-6
E=6×104N/C
mg=F
(F=qE) 
mg=qE
3×10-6×10=q×6×104
By dividing the value of E by Left Hand Side you will get the answer written below
q=5×10-10C
Last Activity: 5 Years ago
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