A charged cork of mass m suspended by a light string is placed in uniform electric field of strength E = ( i^ + j^ ) * 10 5 N/C. If in equilibrium position tension in string is 2mg/ ( 1 + root3), the angle ‘alpha’ with the vertical is :

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Nikhil · Answer

Actually answer is 1st bcoz ( cosa +sina)=1+√3/2 = 1/2+√3/2.. ~cos a =1/2= 60 .. Sina =√3/2=60 .. Actually in this F = MA WILL ACT. AND VECTOR COMPONETA ARE RECUIRED

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A charged cork of mass m suspended by a light string is placed in uniform electric field of strength E = ( i^ + j^ ) * 10 5 N/C. If in equilibrium position tension in string is 2mg/ ( 1 + root3), the angle ‘alpha’ with the vertical is :

A charged cork of mass m suspended by a light string is placed in uniform electric field of strength E = ( i^ + j^ ) * 10⁵N/C. If in equilibrium position tension in string is 2mg/ ( 1 + root3), the angle ‘alpha’ with the vertical is :

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A charged cork of mass m suspended by a light string is placed in uniform electric field of strength E = ( i^ + j^ ) * 10 5 N/C. If in equilibrium position tension in string is 2mg/ ( 1 + root3), the angle ‘alpha’ with the vertical is :

A charged cork of mass m suspended by a light string is placed in uniform electric field of strength E = ( i^ + j^ ) * 105 N/C. If in equilibrium position tension in string is 2mg/ ( 1 + root3), the angle ‘alpha’ with the vertical is :

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A charged cork of mass m suspended by a light string is placed in uniform electric field of strength E = ( i^ + j^ ) * 10⁵N/C. If in equilibrium position tension in string is 2mg/ ( 1 + root3), the angle ‘alpha’ with the vertical is :