Question

The electric flux through a square-shaped area of side 5 cm near a large charged sheet is found to be 3 × 10−5 N · m2 /C when the area is parallel to the plate. Find the charge density on the sheet.

Answer 1

Answer:

$\sigma=2.124\times 10^{-13}C/m^{2}$

Explanation:

Given:

Electric Flux = $3\times10^{-5}N.m^2/C$

Side of sheet = 5cm

Area of the square sheet, A= 5×5 = 25cm²=25×10⁻⁴ m²

Now

the electric flux (Φ) is given as:

$\phi =EA$

where, E = Electric field

or

$E=\frac{\phi}{A}$

substituting the values in the above equation, we get

$E=\frac{3\times10^{-5}N.m^2/C}{25\times 10^{-4}}$

$E=0.012N/C$

Now the charge density (σ) on a sheet is given as:

$\sigma=2\epsilon_oE$

where, $\epsilon_o$ = Permittivity of the free space = 8.85×10⁻¹²

substituting the values in the above equation, we get

$\sigma=2\times 8.85\times10^{-12}\times 0.012$

$\sigma=2.124\times 10^{-13}C/m^{2}$

Answer 2

The charge density on the sheet is 2.12 x 10⁻¹³ C/m².

Electric field

The electric field experienced by the charge is calculated as follows;

Ф = EA

where;

• Ф is flux
• A is area
• E is electric field strength

$E = \frac{\Phi }{A} \\\\E = \frac{3 \times 10^{-5} }{(0.05)^2} \\\\E = 0.012 \ N/C$

Charge density

The charge density on the sheet is calculated as follows;

σ = 2εE

σ = 2 x 8.85 x 10⁻¹² x 0.012

σ =  2.12 x 10⁻¹³ C/m²

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