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

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20,000 N/C

10,000 N/C

40,000 N/C

90,000 N/C

answer is C.

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Detailed Solution

Electric field upwards and charge is negative hence force acts in downwards
$\begin{array}{l} ∴ g^{'} = g + \frac{Eq}{m} = \frac{10 + E \times 10^{- 6}}{1 \times 10^{- 3}}; g^{'} = 10 + (E \times 10^{- 3}) \\ R = \frac{u^{2} \sin 2 θ}{g'} \Rightarrow 2 = \frac{10^{2} \times \sin 90^{0}}{g^{'}} \Rightarrow g^{'} = 50 \\ ∴ g^{'} = 10 + E \times 10^{- 3}; 50 = 10 + E \times 10^{- 3} \\ 40 = E \times 10^{- 3} \Rightarrow E = 40 \times 10^{3} = 40000 N / C \end{array}$