Question

How much work is required to deflect the dipole of charge 3×10^-6C and length 50cm through an angle 60degree in a electric field 1×10^5 N/C.​

Answer 1

Answer:

Torque on dipole, T = P×E×Sin( ¥ )

Where 

P = Dipole Moment = 2qr = q× ( 2 r ) 

Here q = Charge on dipole , 

2r = length of dipole = 2cm = 0.02m

E = Electric Field intensity = 1000000 N/C

¥ = angle between P and E = 600

Now

T = P×E×Sin ( ¥ )

  = q×(2r)×E× Sin ( ¥ )

 = q×2r×E× Sin ( ¥ )

q = 2r×E×Sin(¥)T

= 0.02×100000083×(23)

= 2×1000000×(3)[83)×100×2]

= 100008

= 8×10−4C

= 8×10−4C×100100

= 8×100×[10010−4]C

= 800×10−6 C

= 800 microCoulomb