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How does the electric flux due to a ...

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

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remains same as it does not depend upon shape and size of Gaussian surface
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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))`

  • The electric flux through a closed surface area S enclosing charge Q is phi If the surface area is doubled, then the flux is

    A
    `2phi`
    B
    `phi//2`
    C
    `phi//4`
    D
    `phi`

  • The electric flux through a chlosed surface area S enclosing charge Q is phi . If the surface area is doubled, then the flux is

    A
    `2phi`
    B
    `phi//2`
    C
    `phi//4`
    D
    `phi`

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    A positive charge q is enclosed by a Guassian spherical surface of radius 'a' . If its radius is increased to 4a then the net outward flux will

    Define electric flux. Write its SI unit. A charge q is enclosed by a spherical surface of radius R. If the radius is reduced to half, how would the electric flux through the surface change ?

    The suface integral of electrostatic field vec(E) produced by any sources over any closed surface S enclosing a volume V in vacumm, i.e., total electric flux over the closed S in vacumm is 1//in_(0) times the total charge (Q) contained inside S, i.e, phi_(E) = oint vec(E). vec(ds) = (Q)/(in_(0)) The charge inside S may be point charges or even continous charge distributions. There is no contribution to total electric flux from the charges outside S. Further, the location at Q inside S does not affect the value of surface integral. Read the above passage and answer the following questions : (i) what are the SI unit and dimensions of electric flux ? (ii) A closed surface in vacumm encloses charge -q, + 3q and +5q . Another charge +4q lies outside the surface. What is total electric flux over the surface ? (iii) A point charge q lies inside a spherical of radius r. How will the electric flux be affected if radius of the sphere is doubled ? (iv) What values of life do yo+-earn from the theorem ?

    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.

    The flux of electric field due to these charges through the surface S is

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