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TS EAMCET · Physics · Capacitance

A particle of mass m and charge q is thrown perpendicular to an electric field of intensity E with an initial velocity v. The particle moves a distance x perpendicular to the field and a distance y along the direction of the field. If y=αx2 then the α is given by

  1. A qEm
  2. B qEv2m
  3. C 2qEmv2
  4. D qE2mv2
Verified Solution

Answer & Solution

Correct Answer

(D) qE2mv2

Step-by-step Solution

Detailed explanation

Acceleration of the charged particle is, a=qEm Perpendicular to the field, acceleration is zero. Therefore, x=vt ⇒t=xv Along y direction, using second equation of motion, we have y=0+12at2 ⇒y=12qEmxv2 ⇒y=qEx22mv2 On comparing to given relation, we have…
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