22. A charge of mass 'm' charge 2e' is
released from rest in a uniform electric
field of strength 'E'. The time taken by
it to travel a distance 'd' in the field is
dm
2dm
2dE
1) t=
2)
3)
4
)
2Ee
V dm
Ee
V Ee
me
Answers
L=2m,
L=2m,d=3mm,A=
L=2m,d=3mm,A= 4
L=2m,d=3mm,A= 49π
L=2m,d=3mm,A= 49π
L=2m,d=3mm,A= 49π ×10
L=2m,d=3mm,A= 49π ×10 −6
L=2m,d=3mm,A= 49π ×10 −6 m
L=2m,d=3mm,A= 49π ×10 −6 m 2
L=2m,d=3mm,A= 49π ×10 −6 m 2
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL=
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 4
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .
L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .