Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are charged with equal amount of opposite charges, ±17 µC. The charges on the plates face each other. Find the flux (in N · m2/C) through a circle of radius 3.5 cm between the plates when the normal to the circle makes an angle of 4° with a line perpendicular to the plates. Note that this angle can also be given as 180° + 4°.

Electric flux;

Φ = 30.095 × 10⁴ N.m²/C

Explanation:

We are given;

Charge on plate; q = 17 µC = 17 × 10^(-6) C

Area of the plates; A_p = 180 cm² = 180 × 10^(-4) m²

Angle between the normal of the area and electric field; θ = 4°

Radius;r = 3 cm = 3 × 10^(-2) m = 0.03 m

Permittivity of free space;ε_o = 8.85 × 10^(-12) C²/N.m²

The charge density on the plate is given by the formula;

σ = q/A_p

Thus;

σ = (17 × 10^(-6))/(180 × 10^(-4))

σ = 0.944 × 10^(-3) C/m²

Also, the electric field is given by the formula;

E = σ/ε_o

E = (0.944 × 10^(-3))/(8.85 × 10^(-12))

E = 1.067 × 10^(8) N/C

Now, the formula for electric flux for uniform electric field is given as;

Φ = EAcos θ

Where A = πr² = π × 0.03² = 9π × 10^(-4) m²

Thus;

Φ = 1.067 × 10^(8) × 9π × 10^(-4) × cos 4

Φ = 30.095 × 10⁴ N.m²/C

0.863333

86%

hope this

higher potential

explanation:

because it is in a high speed so it must move into a of high potential