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3 years ago

If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be

Physics

1 answer:

il63 [147K]3 years ago

4 0

Answer:

Explanation:

We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .

∫ E ds = q / ε

Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so

∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .

This also represents total flux coming out of the charge q on all sides .

This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .

If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .

It is so because flux coming out of charge q is q / ε . It does not depend upon surface area enclosing the charge . It depends upon two factors

1 ) charge q and

2 ) the permittivity of medium ε around .

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Answer:

V = 9.682 × 10^(-6) V

Explanation:

Given data

thick = 190 µm

wide = 4.20 mm

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Calculate V

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3 years ago

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2 years ago

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Explanation:

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We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

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where:

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Substituting everything into the equation we get:

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Solving the equation for t, we find the time of flight of the ball:

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Therefore, the horizontal distance covered in a time t is

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And substituting t = 3.59 s, we find

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So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

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