An Oil Drop of mass 0.03 gram carrying a negative charge of 3e is held in equilibrium between two metal plates 1.5 CM apart and across which PD is maintained what is the intensity of electric field and PD between the plate . Given : e=1.6x10^-19, g= 9.8m/s

Dear Student,

m=0.03 gm , q=-3e, d=1.5 cm , let the field be EqE=mg3*1.6*10-19*E=0.03*10-3*9.8E=0.03*10-3*9.83*1.6*10-19=6.125*1014 N/CPD=E*1.5*10-2=6.125*1014 *1.5*10-2=9.1875*1012 V Regards

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