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How does the electric flux due to a poin...

How does the electric flux due to a point charge enclosed by a spherical Gaussian surface altered when its radius is increased?

Text Solution

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The electric flus does not alter at all and remains uncharged.
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How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?

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Knowledge Check

  • A charge q is enclosed by an imaginary Gaussian surface. If radius of surface is increasing at a rate (dr)/(dt)= K , then

    A
    flux linked with surface is increasing at a rate `(dphi)/(dt)=K`
    B
    flux linked with surface is decreasing at a rate `(dphi)/(dt)=-K`
    C
    flux linked with surface is decreasing at a rate `(dphi)/(dt)=(1)/(K)`
    D
    flux linked with surface is `(q)/(epsilon_(0))`

  • A point charge produces an electric flux of -1xx10^(3)Nm^(2)//C to pass through a spherical gaussian surface of radius 10 cm cenred on the charge. What is the value of the point charge?

    A
    `7.85xx10^(-6)C`
    B
    `8.85xx10^(9)C`
    C
    `-8.85xx10^(-9)C`
    D
    `4.425xx10^(-9)C`

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    Which of the following statements is//are correct? a. Electric field calculated by Gauss law is the field due to only those charges which are enclosed inside the Gaussian surface. b. Gauss law is applicable only when there is a symmetrical distribution of charge. c. Electric flux through a closed surface is equal to total flux due to all the charges enclosed within that surface only.

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