Question
A potential difference of $5000$ volt is applied between two parallel plates $5\ cm$ apart. A small oil drop having a charge of $9.6 \time 10-19 C$ falls between the plates. Find (i) electric field intensity between the plates and (ii) the force on the oil drop.

Answer

Given: $V = 5000 volt, d = 5 cm = 5 \times 10^{-2} m$
$q = 9.6 \times 10^{-19} C$
To find:
i. Electric field intensity (E)
ii. Force (F)
Formula:
i. $E =\frac{V}{d} \frac{q}{r}$
ii. $E =\frac{F}{q}$
Calculation: From formula (i),
$E =\frac{F}{q}=10^5 N / C$
From formula (ii)
$ F=E \times q$
$=10^5 \times 9.6 \times 10^{-19}$
$=9.6 \times 10^{-14} N $

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