Electric Field And Field Lines

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A tiny oil droplet carrying a charge of 5 nanocoulomb experiences a force of 2 millinewton in a laboratory field, so what is the field strength?

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Electric field is defined as the force experienced per unit positive charge, expressed as $E = \frac{F}{q}$, a relation emphasised in NCERT Class 12, Chapter 1 (Electric Charges and Fields). The droplet carries $q = 5	imes10^{-9}$ C and feels a force $F = 2	imes10^{-3}$ N. Dividing gives $E = \frac{2	imes10^{-3}}{5	imes10^{-9}}$. Working the powers of ten, $\frac{2}{5} = 0.4$ and $\frac{10^{-3}}{10^{-9}} = 10^{6}$, so $E = 0.4	imes10^{6} = 4	imes10^{5}$ N/C. The value $2.5	imes10^5$ N/C is wrong because it inverts the ratio, dividing charge by force. The value $1	imes10^6$ N/C is wrong because it rounds $0.4$ up to $1$ without justification. The value $1	imes10^4$ N/C is wrong because it mishandles the exponent difference by two orders. This laboratory setup echoes Millikan's oil-drop experiment where droplets are balanced in fields. A magnitude check confirms nano-coulomb charges in milli-newton forces need fields of order $10^5$ N/C, matching the answer's exponent and sign.

Electric Field And Field Lines

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