First slide

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 - $1 μ C$ 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 $m / s^{2}$ )

Moderate

A

20,000 N/C
B

10,000 N/C
C

40,000 N/C
D

90,000 N/C

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 $g + \frac{qE}{m}$
$R = \frac{V^{2} \sin 2 θ}{g + \frac{qE}{m}} = \frac{V^{2}}{g + \frac{qE}{m}} (θ = 45^{o})$
$R = \frac{100}{10 + \frac{10^{- 6} E}{10^{- 3}}} = 10 + 10^{- 3} E = 50$
$E = 40 \times 10^{3} N / C$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App