A particular oil drop is suspended in a field with an intensity of 2.19 x 103N/C. If the apparent weight of the oil drop equal 1.4 x 10-15N. Calculate the charge of the oil drop in C?
E=Fq ⟹ q=FE,E=\frac Fq\implies q=\frac FE,E=qF⟹q=EF,
q=1.4⋅10−152.19⋅103=0.64⋅10−18 C.q=\frac{1.4\cdot10^{-15}}{2.19\cdot 10^3}=0.64\cdot 10^{-18}~\text C.q=2.19⋅1031.4⋅10−15=0.64⋅10−18 C.
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