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2. Electric Potential and Capacitance
easy
An electron enters between two horizontal plates separated by $2\,mm$ and having a potential difference of $1000\,V$. The force on electron is
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
(b) Force on electron $F = QE$ $ = Q\,\left( {\frac{V}{d}} \right)$
$⇒$ $F = (1.6 \times {10^{ – 19}})\,\left( {\frac{{1000}}{{2 \times {{10}^{ – 3}}}}} \right) = 8 \times {10^{ – 14}}\,N$
Standard 12
Physics
