Question

Part A 6 x 10"9C-the charge of the electron. For this he en the Nobel Prize. (Figure 1) ure If a drop of mass 1.4 x 10" kg remains stationary in an electric field of 1.715 x 10% N/C, what is the charge of this drop? Express your answer to two significant figures and include the appropriate units. 1 of 1 HA Comizer Value Units Oil drop Submit Request Answer

Answer 1

We know,

q.E = m.g

here, q = charge of the drop.

E = electric field= 1.715×10^5 N/C

m = mass of the drop = 1.4×10^(-14) kg

g = gravitational acceleration = 9.8 m/s²

Now, q×1.715×10^(5) = 1.4×10^(-14)×9.8

=> q = 8×10^(-19) C

The charge of the drop will be 8×10^(-19) C.

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