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If a charge is enclosed by the surface o...

If a charge is enclosed by the surface of the sphere then total flux emitted from the surface will be:

A

(the charge enclosed by surface)/`epsi_(0)`

B

(charge enclosed by surface) `epsi_(0)`

C

(charge enclosed by surface)/`4pi epsi_(0)`

D

0

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The correct Answer is:
To solve the problem of determining the total electric flux emitted from the surface of a sphere when a charge is enclosed within it, we can use Gauss's Law. Here’s the step-by-step solution: ### Step 1: Understand Gauss's Law Gauss's Law states that the total electric flux (Φ) through a closed surface is directly proportional to the charge (q) enclosed within that surface. Mathematically, it is expressed as: \[ \Phi = \frac{q}{\epsilon_0} \] where: - \( \Phi \) is the electric flux, - \( q \) is the charge enclosed, - \( \epsilon_0 \) is the permittivity of free space. ### Step 2: Identify the Charge Enclosed In this problem, we are given that there is a charge \( +q \) enclosed by the spherical surface. ### Step 3: Apply Gauss's Law According to Gauss's Law, we can substitute the charge \( q \) into the formula for electric flux: \[ \Phi = \frac{q}{\epsilon_0} \] ### Step 4: Conclusion From our application of Gauss's Law, we find that the total electric flux emitted from the surface of the sphere is: \[ \Phi = \frac{q}{\epsilon_0} \] Thus, the correct answer is that the total flux emitted from the surface will be the charge enclosed by the surface divided by \( \epsilon_0 \). ### Final Answer The total flux emitted from the surface will be: \[ \frac{q}{\epsilon_0} \]
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AAKASH SERIES-ELECTRIC CHARGES AND FIELDS-Exercise-I
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  6. Gauss.s law is true only if force due to a charge varies as:

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  11. A thin conducting spherical shell of radius R has charge Q spread unif...

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  14. An electric dipole is plced at an angle of 30^(@) to a nonuniform elec...

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  16. An electric dipole is kept in non-uniform electric field. It experienc...

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  17. Match List-I with List-II

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