1.

A cube of edge length 2.00m is placed in a region and which the electric field is given by E=(20.00-28.00z) k N/C. Determine the net charge contained within the cube

Answer»

Given :

Edge LENGTH of the given cube = 2 m

The electric field in that region in which  the cube is placed  is given by :

E= ( 20.00 - 28.00z ) \frac{kN}{C}

To  Find :

The amount of net charge contained within the cube = ?

Solution:

Here the electric field along x and y direction is ZERO since E is acting along only z direction .

In this cube we have taken two surfaces as shown in the attached fig .

So, the net electric flux = Flux from 2ND surface - flux from 1st surface

( ∴ flux going out is taken as positive and flux coming in is taken as negative)

So, \Phi_{Enet} = \Phi_2 - \Phi_1

Here \Phi_2 is going out so it is +ve and \Phi_1 is coming in so it is -ve .

So, \Phi_{Enet} = E(z).A - E(z-a).A

                =(20 - 28z).A -[20-28(z-a)].A\\

                =A[20-28z-20+28z-28a]

                =-A.(28a)

Here A is area of one face and a is edge length , since it is given that edge length is 2 m .

So, a = 2 m

And A = (2)2 m^2 = 4 m^2

Hence , \Phi{Enet}= 28 \times 2 \times 4

                         =224 NC^{-1}m{-2}

By Gauss law we have ;

Flux = \frac{charge \hspace3 enclosed \hspace3 inside \hspace3 the \hspace3 cube }{\epsilon_0}

Or, 224 = \frac{<klux>Q</klux>}{\epsilon_0}     ( here Q is charge enclosed )

Or, Q = 224 \times 8.85 \times 10^{-12} C

          =1.982 \times 10^{-9} C

Hence , the net charge contained within the cube is 1.982 \times 10^{-9} C .


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