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The electric field in a region is given ...

The electric field in a region is given by `vecE=3/5 E_0veci+4/5E_0vecj` with `E_0=2.0 xx 10^3 N C^(-1)`. Find the flux of this field through a recatngular surface of area `0.2m^2` parallel to the y-z plane.

Text Solution

AI Generated Solution

To find the electric flux through a rectangular surface of area \(0.2 \, \text{m}^2\) that is parallel to the y-z plane, we can follow these steps: ### Step 1: Write down the electric field vector The electric field vector is given by: \[ \vec{E} = \frac{3}{5} E_0 \hat{i} + \frac{4}{5} E_0 \hat{j} \] where \(E_0 = 2.0 \times 10^3 \, \text{N/C}\). ...
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Knowledge Check

  • The electric field in a region is given by E = 3/5 E_0hati +4/5E_0j with E_0 = 2.0 x 10^3 N//C . Find the flux of this field through a rectangular surface of area 0.2 m^2 parallel to the y-z plane.

    A
    `24Nm^2//C`
    B
    `240Nm^2//C`
    C
    `440Nm^2//C`
    D
    `44Nm^2//C`
  • Consider a uniform electric field E = 3 xx 10^(3) hat(i)N C^(-1) . What is te flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ?

    A
    `10 N C^(-1) m^(2)`
    B
    `20 N C^(-1) m^(2)`
    C
    `30 N C^(-1) m^(2)`
    D
    `40 N C^(-1) m^(2)`
  • An electric field given by E=2E_(0)veci+3E_(0) vecj-5E_(0)veck exists in a cerrtain region, where E_(0)=100N//C . How much flux will pass through a rectangular surface of area 0.2 m^(2) , placed in this region, in such away that it is parallel to Y-axis?

    A
    `20Nm^(2)//C`
    B
    `40Nm^(2)//C`
    C
    `60Nm^(2)//C`
    D
    `80Nm^(2)//C`
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