Motion of charged particle in uniform electric field
Question

# An electric field is acting vertically upwards. A small body of mass 1 gm and charge  -$1\mu C$  is projected with a velocity 10 m/s at an angle ${45}^{0}$ with horizontal. Its horizontal range is 2m then the intensity of electric field is : (g = 10 $\mathrm{m}/{\mathrm{s}}^{2}$)

Moderate
Solution

## As charge is negative and the electric field is vertically upwards the total force acting on the charge is vertically downwards.  Hence the net acceleration downwards is $\mathrm{g}+\frac{\mathrm{qE}}{\mathrm{m}}$$\mathrm{R}=\frac{{\mathrm{V}}^{2}\mathrm{sin}2\mathrm{\theta }}{\mathrm{g}+\frac{\mathrm{qE}}{\mathrm{m}}}=\frac{{\mathrm{V}}^{2}}{\mathrm{g}+\frac{\mathrm{qE}}{\mathrm{m}}}\left(\mathrm{\theta }={45}^{\mathrm{o}}\right)$$\mathrm{R}=\frac{100}{10+\frac{{10}^{-6}\mathrm{E}}{{10}^{-3}}}=10+{10}^{-3}\mathrm{E}=50$$\mathrm{E}=40×{10}^{3}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{N}/\mathrm{C}$

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