PhysicsEasy46×since 2002Q6953A charged oil drop is suspended in a uniform field of 3×1043 \times {10^4}3×104 v/mv/mv/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge =9.9×10−15 kg= 9.9 \times {10^{ - 15}}\,\,kg=9.9×10−15kg and g=10 m/s2g = 10\,m/{s^2}g=10m/s2)A1.6×10−18 C1.6 \times {10^{ - 18}}\,C1.6×10−18CB3.2×10−18 C3.2 \times {10^{ - 18}}\,C3.2×10−18CC3.3×10−18 C3.3 \times {10^{ - 18}}\,C3.3×10−18CD4.8×10−18 C4.8 \times {10^{ - 18}}\,C4.8×10−18CCheck answerSkip