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sweet [91]
3 years ago
6

A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe

r face carries +95.0 nC/m2 while the lower face carries -25.0 nC/ m2. What is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? (ε0 = 8.85 × 10-12 C2/N · m2) A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The upper face carries +95.0 nC/m2 while the lower face carries -25.0 nC/ m2. What is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? ( = 8.85 × 10-12 C2/N · m2) 0.00 N/C 6.78 × 103 N/C 7.91 × 103 N/C 1.36 × 104 N/C 3.95 × 103 N/C
Physics
1 answer:
Nonamiya [84]3 years ago
4 0

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

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\large \boxed{42\, \mu \text{C}}$

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\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

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