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
B
5.0×1011N/C
C
5.5×1011N/C
D
56 N/C

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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