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3 years ago

12

The electric field in a certain region of Earth’s atmo- sphere is directed vertically down. At an altitude of 300 m the field ha

s magnitude 60.0 N/C; at an altitude of 200 m, the magnitude is 100 N/C. Find the net amount of charge contained in a cube 100 m on edge, with horizontal faces at altitudes of 200 and 300 m

2 answers:

I am Lyosha [343]3 years ago

6 0

Answer:

Net charge contained in the cubeq= 3.536×10^-6C

Explanation:

Formular for total flux in a cube is given as:

Total flux= E300Acos(180) + E200Acos(0)

Where A is crossectional area

Total flux= A(E200-E300)

Total flux= q/Eo

q= Eo×total flux

q=(8.84×10^-12)×(100)^2×(100-60)

q= 3.536×10^-6C

Alisiya [41]3 years ago

3 0

Answer:

$q = 3.54*10^{-6}C$

Explanation:

Given that:

A = 100m

( $E_l$ ) = 100

( $E_v$ ) = 60

E₀(constant) = 8.85 * 10⁻¹²

The net amount of charge can be found using the expression:

= $\frac{q}{E_0} = \theta$

= $q= \theta E_0$ where θ =0; A(El-Ev)

= $q= E_0A(E_l-E_v)$

= $q=(8.85*10^{-12}*(100)^2*(100-60)$

= $q=(8.85*10^{-12}*(100)^2*(40)$

= $q=(8.85*10^{-12}*(400,000)$

= $q = 3.54*10^{-6}C$

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