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Question

Find out the surface charge density at the intersection of point x=3mplane and x-axis in the region of uniform linear charge density of 8nC/m lying along the z-axis in free space.


  1. 47.88C/m

  2. 0.07nCm-2

  3. 0.424nCm-2

  4. 4.0nCm-2

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Solution

The correct option is C

0.424nCm-2


Step 1. Given data.

Intersection point, x=3m,

Uniform line charge that is linear charge density, λ=8nC/m,

Assume coulomb constant k=14πε0 (where ε0 is the permittivity of free space),

ε0=8.85×10-12Fm-1 and π=3.14.

Step 2. The formula for Electric field due to wire,

Electric field due to wire is given by E=2kλx1,

Where, E is the Electric field due to wire, k is the coulomb constant, λ is the linear charge density and x is the perpendicular distance of the point from the line charge.

Step 3. The formula for Electric field with surface charge density,

The electric field with surface charge density E=σε2,

Where, E is the electric field with surface charge density, σ is the surface charge density and ε0 is the permittivity of free space,

Step 4. Calculating surface charge density,

Now, by equating the above equation 1 and equation 2 we get,

2kλx=σε214πελx=σεσ=2λε04πxεσ=λ2πxσ=8×10-92×3.14×3Cm-2σ=8×10-918.84Cm-2σ=0.424×10-9Cm-2σ=0.424nCm-2

Therefore, the surface charge density, σ is0.424nCm-2.

Hence, the correct option is C.


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