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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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3 years ago

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2 years ago

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2 years ago

A conducting sphere contains positive charge distributed uniformly over its surface. Which statements about the potential due to

stira [4]

Answer:

a. FALSE

b. FALSE

c. TRUTH

d. FALSE

e. FALSE

Explanation:

To determine which statements are truth or false you focus in the following formula, for the electric potential generated by a conducting sphere:

$V=\frac{Q}{4\pi \epsilon_o R}$ inside the sphere

$V'=\frac{Q}{4\pi \epsilon_o r}$ for r > R (outside the sphere)

R: radius of the sphere

ε0: dielectric permittivity of vacuum

Q: charge of the sphere

As you can notice, inside the sphere the potential is constant. Inside the sphere, the potential is the same. Outside the surface the potential decreases as 1/r, being r the distance to the center of the sphere.

Hence, you can conclude:

a. The potential at the center of the sphere is zero. FALSE

b.The potential is lowest, but not zero, at the center of the sphere. FALSE

c. The potential at the center of the sphere is the same as the potential at the surface. TRUTH

d. The potential at the center is the same as the potential at infinity. FALSE

e. The potential at the surface is higher than the potential at the center. FALSE

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3 years ago

(BRAINLIEST W/ WORK SHOWN!!!) Help with Physics question?

VashaNatasha [74]

Answer:

$\boxed {3.43 x 10^{3}}$

Explanation:

We know that speed is defined as distance moved per unit time hence expressed as $v=\frac {d}{t}$ where v is speed in m/s, d is distance in m and t is time in seconds. Making d the subject of the above formula then

$d=vt$

Substituting 343 m/s for d and 10 s for t then

$d= 343\times10= 3430= 3.43 x 10^{3}$

Therefore, the distance between speaker and deter is $\boxed {3.43 x 10^{3}}$

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2 years ago