Physics, asked by ShindePrathmesh, 2 months ago



Electric charge is uniformly distributed along a long straight
wire of radius 1mm. The charge per cm length of the wire is
Q coulomb. Another cylindrical surface of radius 50 cm and
length 1m symmetrically encloses the wire as shown in the
figure. The total electric flux passing through the cylindrical
surface is
[MP PET 2001; Kerala PET 2011]
Q
(a)
EO
ttt
1000
(b)
1
EO
1m
(c)
100
(TE)
100Q
(TTE )
(d)
+
+ 50cm
+​

Answers

Answered by anubhabkumar2020
0

Answer:

Answer:

Answer:

Answer: L=2m,

Answer: L=2m,d=3mm,A=

Answer: L=2m,d=3mm,A= 4

Answer: L=2m,d=3mm,A= 49π

Answer: L=2m,d=3mm,A= 49π

Answer: L=2m,d=3mm,A= 49π ×10

Answer: L=2m,d=3mm,A= 49π ×10 −6

Answer: L=2m,d=3mm,A= 49π ×10 −6 m

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL=

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 4

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5

Answer: 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

Answer: 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

Answer: 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 .

Answer: 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 .

Answer: 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 .

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