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Q.

An electric field is acting vertically upwards, A small body of mass 1gm and charge $- 1 μC$ is projected with a velocity 10m/s at an angle 45° with horizontal. Its horizontal range is ‘2m’ then the intensity of electric field is (g =10m/s²)

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a

10,000 N/C

b

90,000 N/C

c

20,000 N/C

d

40,000 N/C

answer is C.

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Detailed Solution

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Range $R = \frac{u^{2} \sin 2 θ}{g + Eq / m} = \frac{{(10)}^{2} \times \sin (2 \times 45)}{10 + E \times (1 \times 10^{- 6}) / 1 \times 10^{- 3}}$

$2 = \frac{10^{2} \times 1}{10 + \frac{E \times 10^{- 6}}{10^{- 3}}}$ $\Rightarrow E = 40, 000 N / C$

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An electric field is acting vertically upwards, A small body of mass 1gm and charge −1μC is projected with a velocity 10m/s at an angle 45° with horizontal. Its horizontal range is ‘2m’ then the intensity of electric field is (g =10m/s2)