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Question

An electron experience a force equal to its weight when placed in an electric field. The intensity of the field will be:


A
1.7×1011N/C
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B
5.0×1011N/C
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C
5.5×1011N/C
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D
56 N/C
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Solution

The correct option is B $$5.5\times 10^{-11}N/C$$
Force acting on electron is given by:
$$ F=qE$$
$$\Rightarrow E=\frac{F}{q}=\frac{mg}{e}$$
$$=\frac{9\times 10^{-31}\times 9.8}{1.6\times 10^{-19}}$$
$$=5.5\times 10^{-11}N/C$$

Physics

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