An electric field is acting vertically upwards- A small body of mass 1 gm and charge -1- C is projected with a velocity 10 m-s at an angle 45- with horizontal- Its horizontal range is 2m then the intensity of electric field is- -g - 10 m-s-2-20-000 N-C10-000 N-C40-000 N-C 90-000 N-C

The correct option is C 40,000 N/C R=u^2 sin 2 θ/g+a=u^2 sin 2 × 45/g+Eg/m 2=10^2 × sin 90/10+E × 10^ 6/10^ 3 10+10^ 3E=50 E=40 × 10^3=40,000 N/C ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Range on an Incline

An electric f...

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

20,000 N/C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10,000 N/C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40,000 N/C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

90,000 N/C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 40,000 N/C
$R = \frac{u^{2} s i n 2 θ}{g + a} = \frac{u^{2} s i n 2 \times 45}{g + \frac{E g}{m}}$
$2 = \frac{10^{2} \times s i n 90}{10 + \frac{E \times 10^{- 6}}{10^{- 3}}}$
$10 + 10^{- 3} E = 50$
$E = 40 \times 10^{3} = 40, 000 N / C$

Suggest Corrections

Similar questions

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