MCQ
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
    $5.0 \times {10^{ - 11}}\,N/C$
  • $5.5 \times {10^{ - 11}}\,N/C$
  • D
    $56 N/C$

Answer

Correct option: C.
$5.5 \times {10^{ - 11}}\,N/C$
c
(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$

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