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The electric field in a region is given ...

The electric field in a region is given by `E = (E_0x)/lhati`. Find the charge contained inside a cubical 1 volume bounded by the surfaces `x=0, x=a, y=0, yh=, z=0` and z=a. Take `E_0=5xx10^3N//C` l=2cm and a=1m.

A

`1.1 xx 10^(-12)C`

B

`2.2 xx 10^(-12)C`

C

`4.4 xx 10^(-12)C`

D

`5.5 xx 10^(-12)C`

Text Solution

Verified by Experts

The correct Answer is:
B
Given `E=(E_(0))/(l)vec(i).l=2cm,a=1cm,E_(0)=5xx10^(3)NC^(-1)`
We see that flux passes mainly through surface area `ABDC` and EFGH. As the AEFB and CHGD are parallel to the flux again ABDC`=0` the flux only passes through the surface area `EFGHE=0`
Flux `=E_(0)(a)/(l)"area"=5xx10^(3)xx(a)/(l)xxa^(2)`
`=5xx10^(3)xx(a^(3))/(l)`
`=5xx10^(3)xx((0.01)^(3))/(2xx10^(-2))=2.5xx10^(-1)`
so `q=epsilon_(0)` flux
`=8.85xx10^(12)xx2.5xx10^(-1)`
`=22.125xx10^(-13)=2.2125xx10^(-12)C`
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