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1 year ago

5

A flat surface of area 3.20m² is rotated in a uniform electric field of magnitude E=6.20 × 10⁵N . m²/C . Determine the electric

flux through this area (a) when the electric field is perpendicular to the surface

1 answer:

lana66690 [7]1 year ago

8 0

The electric flux is 0 N.m²/C.

We need to know about electric flux to solve this problem. Electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow. It can be determined as

Φ = E . S = E . Scosθ

where Φ is electric flux, E is electric field, θ is angle of surface and S is surface area.

From the question above, we know that

E = 6.20 × 10⁵ N/C

θ = 90⁰

S = 3.20 m²

By substituting the following parameter, we get

Φ = E . Scosθ

Φ = 6.20 × 10⁵ . 3.20cos(90⁰)

Φ = 0 N.m²/C

Hence, the electric flux is 0 N.m²/C

For more on electric flux at: brainly.com/question/26289097

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What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?

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Answer:

$F=1.26*10^{-3}N$

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Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

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We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges( $q_1,q_2$ ) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

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4.1 A steel spur pinion has a pitch of 5 teeth/in, 20 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of

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Answer:

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Pitch (P) = 5 teeth/in

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Power (H) = 30 hp

Teeth on gear ( $N_G$ ) = 50

Teeth on pinion ( $N_p$ ) = 20

Face width (F) = 1 in

Let us first determine the diameter (d) of the pinion.

Diameter (d) = $\frac{N}{P}$

= $\frac{20}{5}$

= 4 in

From the values of Lewis Form Factor Y for ( $n_p$ ) = 20 ; at 20°

Y = 0.321

To find the velocity (V); we use the formula:

$V = \frac{\pi d n_p}{12}$

$V = \frac{\pi *4*2000}{12}$

V = 2094.40 ft/min

For cut or milled profile; the velocity factor $(K_v)$ can be determined as follows:

$K_v = \frac{2000+V}{2000}$

$K_v = \frac{2000+2094.40}{2000}$

= 2.0472

However, there is need to get the value of the tangential load $(W^t)$ , in order to achieve that, we have the following expression

$W^t=\frac{T}{\frac{d}{2} }$

$W^t = \frac{63025*H}{\frac{n_pd}{2}}$

$W^t = \frac{63025*30}{2000*\frac{4}{2}}$

$W^t = 472.69 lbf$

Finally, the bending stress is calculated via the formula:

$\sigma = \frac{K_vW^tp}{FY}$

$\sigma = \frac{2.0472*472.69*5}{1*0.321}$

$\sigma = 15073.07 psi$

$\sigma =$ 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi

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