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1. Electric Charges and Fields
easy
An electron experiences a force equal to its weight when placed in an electric field. The intensity of the field will be
A
$1.7 \times {10^{ - 11}}\,N/C$
B
$5.0 \times {10^{ - 11}}\,N/C$
C
$5.5 \times {10^{ - 11}}\,N/C$
D
$56 N/C$
Solution
(c) $E = \frac{F}{q} = \frac{{mg}}{e}$ $ = \frac{{9 \times {{10}^{ – 31}} \times 9.8}}{{1.6 \times {{10}^{ – 19}}}} = 5.5 \times {10^{ – 11}}N/C$
Standard 12
Physics
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