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 ${45}^{\circ }$ with horizontal. Its horizontal range is 2m then the intensity of electric field is: (g = 10 $m/s^2$)

• A. 20,000 N/C
• B. 10,000 N/C
• C. 40,000 N/C
• D. 90,000 N/C

Answer 1

The correct option is C 40,000 N/C

$R=\frac{u^{2}sin2\theta }{g+a}=\frac{u^{2}sin2\times 45}{g+\frac{Eg}{m}}$
$2=\frac{{10}^2\times sin90}{10+\frac{E\times {10}^{-6}}{{10}^{-3}}}$
$10+{10}^{-3}E=50$
$E=40\times {10}^3=40,000N/C$