Question

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​

Answer 1

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 .