Question

03. A hollow cylindrical box of length 1 m and area of cross section 25cm’ is as shown in figure. The electric field in the region is
given by È = 50.x. I, where E is in N/C and vis in meter. Find. (a) Net Flux through the cylinder. (b) Charge enclosed by the
cylinder
1m
1m​

Answer 1

Explanation:

Area of the cross-section of the cylinder A=25×10

−4

m

−2

Electric field at the face nearer to the origin E

1

=20×0.5=10 N/C

Thus flux entering through this face ϕ

1

=E

1

A=10×25×10

−4

=2.5×10

−2

Wb

Distance of second face from origin = 1+0.5=1.5 m

Electric field at the face away from the origin E

2

=20×(1.5)=30 N/C

Thus flux leaving out through this face ϕ

2

=E

2

A=30×25×10

−4

=7.5×10

−2

Wb

(i) : Net flux through the cylinder ϕ=(7.5−2.5)×10

−2

=5×10

−2

Wb

(ii) : Charge enclosed in the cylinder Q=ϕϵ

o

∴ Q=5×10

−2

×8.85×10

−12

=4.42×10

−13

C

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