Electrostatics> A particle of mass 2kg and charge 1mC is ...

A particle of mass 2kg and charge 1mC is projected vertically with a velocity 10m/s. THere is a uniform horizontal electric field of 10^4 N/C. Find the horizontal range, the time of flight and the mazimum height attained by the particle

Atharva thodge , 7 Years ago

Grade 11

2 Answers

Khimraj

Last Activity: 7 Years ago

acceleration in horizontal direction is given by qE/m = 5m/s²

time of flight will be given by

0 = 10T – ½gT²

T = 2 sec

maximum height

H = 10*1 + ½*10*1²

H = 15m.

Horizontal range = ½*5*2² = 10m.

Hope it clears.

Approved

S V Naga Santoshi

Last Activity: 5 Years ago

Acceleration in horizontal direction is qE/m

Since F=qE in this case...so by substituting the values a=5m/s^2..

To find time of flight we know that ..

S=ut+1/2at^2 i.e, at 0=10T -1/2×10×T^2 then we will get T=2seconds...

Maximum height H :

V^2-U^2 =2gH

-100 = 2× (-10) ×H ..

[since body moving upward direction g is negative and at maximum height v=0]

H = 5m …

Thanking you for giving an opportunity...

ifthere is any mistake please forgive me..

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Electrostatics> A particle of mass 2kg and charge 1mC is ...

A particle of mass 2kg and charge 1mC is projected vertically with a velocity 10m/s. THere is a uniform horizontal electric field of 10^4 N/C. Find the horizontal range, the time of flight and the mazimum height attained by the particle

Atharva thodge , 7 Years ago

Khimraj

S V Naga Santoshi

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