First slide

Electric field

Question

An electron enters between two horizontal plates separated by 2mm and having a potential difference of 1000 V. The force on electron is

Easy

A

$8 \times 10^{- 12} N$
B

$8 \times 10^{- 14} N$
C

$8 \times 10^{9} N$
D

$8 \times 10^{14} N$

Solution

Force on electron $F = QE = Q (\frac{V}{d})$
$\Rightarrow F = (1 .6 \times 10^{- 19}) (\frac{1000}{2 \times 10^{- 3}}) = 8 \times 10^{- 14} N$

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