Q2)
shuldn't it be like this
F=qE ( due to electric field [downwards] )
T + qE = 2qE [ qEl = mv2/2 →mv2/l= 2qE ]
T = qE ??
1. Two dipoles of dipole moments p each r placed on points A(a,0) and B(-a,0) as shown in figure.How much work is done in rotating both d dipoles with 90° angle in clockwise direction?
2.A block of mass m and charge q is connected to a point O with help of an inextensible string. The system is on a horizontal table. An electric field is switched on in direction perpendicular to string. What will be tension in string when it become parallel to electric field?
Q1)
i thnk u missed to mention Electric field direction. since u have not mentioned
I'm assuming that direction of E is along the vertical axis ( that is equitorial axis to dipole )
hence angle between the dipole and Electric field E is initially 90° which is considered as zero energy position τ=pEsinθ
let dW is the small amount of workdone in turning the dipole through a small angle dθ.
→ dW = τdθ
→ dW = pEsinθdθ
here we r to rotate 90° clockwise which means in the final position of dipole axis makes 0° with electric lines of force hence we r rotating dipole from zero enrgy position
( 90° ) to 0° position
hence W =
→ W = - pE
Q2)
shuldn't it be like this
F=qE ( due to electric field [downwards] )
T + qE = 2qE [ qEl = mv2/2 →mv2/l= 2qE ]
T = qE ??
2.
@manmay bhaiya it is given that system is kept in horizontal table.
ΔK.E= qEl = mv2/2
=> mv2/l= 2qE
let T be the tension at final position
T - qE=2qE
T= 3qE
arey haan sry for that. Pata nahin kya hogaya tha mujhe question thik se nahin dekha kyun pata nahin.[3]
Thnks for pointing that out.....[1]
but T + qE i guess that shuld be
as qE is towards O and T also [7]