Question

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

• A. $20000N/C$
• B. $10000N/C$
• C. $40000N/C$
• D. $90000N/C$

Answer 1

The correct option is C $40000N/C$

The horizontal range of a projectile is given by $R=\frac{v^2\sin 2\theta }{g}$

When the body is moving in the presence of electric and gravitational field, the body experiences both the weight and electric force, which can be called as apparent weight.

Electric force acting on the negatively charged body is in downward direction because the electric field is in upward direction.

$W_{\text{app}}=W+qE=mg+qE$

Thus, the body feels a different $g$ given by $g_{\text{app}}=g+\frac{qE}{m}$

Range of the projectile, in the presence of electric field is $R=\frac{v^2\sin 2\theta }{g+\frac{qE}{m}}$

Given that $R=2\text{m}$, $\theta ={45}^{\circ }$ and $v=10\text{m/s}$

Thus, $g+\frac{qE}{m}=\frac{v^2\sin 2\theta }{R}=\frac{100}{2}=50$

Using $g=10\text{m/s}$, we have $E=40\frac{m}{q}=40\times \frac{1\times {10}^{-3}}{1\times {10}^{-6}}=40000\text{N/C}$