A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total flux for the surface of the cylinder is given by 

(1) 2πR2E

(2) πR2/E

(3) (πR2πR)/E 

(4) Zero

Subtopic:  Gauss's Law |
 71%
From NCERT
PMT - 1975
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Electric field at a point varies as r0 for

(1) An electric dipole

(2) A point charge

(3) A plane infinite sheet of charge

(4) A line charge of infinite length

Subtopic:  Electric Field |
 69%
From NCERT
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Total electric flux coming out of a unit positive charge put in air is 

(1) ε0

(2) ε01

(3) (4pε0)1

(4) 4πε0 

Subtopic:  Gauss's Law |
 80%
From NCERT
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A cube of side l is placed in a uniform field E, where E=Ei^. The net electric flux through the cube is

(1) Zero

(2) l2E

(3) 4l2E

(4) 6l2E

Subtopic:  Gauss's Law |
 74%
From NCERT
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Eight dipoles of charges of magnitude \((e)\) are placed inside a cube. The total electric flux coming out of the cube will be: 
1. \(\frac{8e}{\epsilon _{0}}\)
2. \(\frac{16e}{\epsilon _{0}}\)
3. \(\frac{e}{\epsilon _{0}}\)
4. zero

Subtopic:  Electric Dipole |
 76%
From NCERT
PMT - 1998
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A charge q is placed at the centre of the open end of the cylindrical vessel. The flux of the electric field through the surface of the vessel is 

(1) Zero

(2) qε0

(3) q2ε0

(4) 2qε0 

Subtopic:  Gauss's Law |
From NCERT
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Electric charge is uniformly distributed along a long straight wire of radius \(1\) mm. The charge per cm length of the wire is \(Q\) coulomb. Another cylindrical surface of radius \(50\) cm and length \(1\) m symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is:

                

1. \(\dfrac{Q}{\varepsilon _{0}}\) 2. \(\dfrac{100Q}{\varepsilon _{0}}\)
3. \(\dfrac{10Q}{\pi\varepsilon _{0}}\) 4. \(\dfrac{100Q}{\pi\varepsilon _{0}}\)
Subtopic:  Gauss's Law |
 64%
From NCERT
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The S.I. unit of electric flux is 

(1) Weber

(2) Newton per coulomb

(3) Volt × metre

(4) Joule per coulomb

Subtopic:  Gauss's Law |
 52%
From NCERT
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Shown below is a distribution of charges. The flux of electric field due to these charges through the surface S is 

(1) 3q/ε0

(2) 2q/ε0

(3) q/ε0 

(4) Zero

Subtopic:  Gauss's Law |
 83%
From NCERT
AIIMS - 2003
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Consider the charge configuration and spherical Gaussian surface as shown in the figure. While calculating the flux of the electric field over the spherical surface, the electric field will be due to: 

(1) q2 only

(2) Only the positive charges

(3) All the charges

(4) +q1 and – q1 only

Subtopic:  Gauss's Law |
 61%
From NCERT
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