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just olya [345]

3 years ago

7

A single point charge q is located at the center of both an imaginary cube and an imaginary sphere. How does the electric flux t

hrough the surface of the cube compare to that through the surface of the sphere?

1 answer:

saveliy_v [14]3 years ago

7 0

Answer:

Explanation:

Electric flux is defined as the flow of electric field intensity through a given surface.

Mathematically:

$\phi=E.A.cos\ \theta$

where:

E = electric field

A = area

θ = angle between the area vector and the electric field

Electric flux through the surface of a sphere will be uniform throughout the surface area due to a charge at the center of the sphere. The distance of the surface from the center is always at a constant distance of radius of the sphere.

Electric flux through the surface of a cube will be varying as the surface area is at a varying distance from the center of the cube. The distance of the surface from the center is not at a uniform distance from the center of the cube and so the projection of solid angle changes.

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A is the correct answer !!!

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Answer:

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Explanation:

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Using formula of force

$F_{net}=F_{in}-F_{out}$

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A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

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Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

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We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

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$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

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that value is your total time

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Explanation:

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