According to Gauss’ Theorem, electric field of an infinitely long straight wire is proportional to 

(1) r

(2) 1r2

(3) 1r3

(4) 1r

Subtopic:  Electric Field | Gauss's Law |
 74%
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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 |
 65%
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 |
 53%
From NCERT
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If the electric flux entering and leaving an enclosed surface respectively is φ1 and φ2, the electric charge inside the surface will be: 

(1) (φ1+φ2)ε0

(2) (φ2φ1)ε0

(3) (φ1+φ2)/ε0

(4) (φ2φ1)/ε0 

Subtopic:  Gauss's Law |
 64%
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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The electric flux for Gaussian surface A that encloses the charged particles in free space is (given q1 = –14 nC, q2 = 78.85 nC, q3 = – 56 nC) 

(1) 103 Nm2 C–1

(2) 103 CN-1 m–2

(3) 6.32 × 103 Nm2 C–1

(4) 6.32 × 103 CN-1 m–2

Subtopic:  Gauss's Law |
 61%
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The electric intensity due to an infinite cylinder of radius R and having charge q per unit length at a distance r(r > R) from its axis is 

(1) Directly proportional to r2

(2) Directly proportional to r3

(3) Inversely proportional to r

(4) Inversely proportional to r2

Subtopic:  Electric Field |
 66%
From NCERT
PMT - 1993
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Two equal negative charges of charge \(-q\) are fixed at the points \((0,a)\) and \((0,-a)\) on the \(Y\text-\)axis. A positive charge \(Q\) is released from rest at the point \((2a,0)\) on the \(X\text-\)axis. The charge \(Q\) will:
1. execute simple harmonic motion about the origin.
2. move to the origin and remain at rest.
3. move to infinity.
4. execute oscillatory but not simple harmonic motion.
Subtopic:  Coulomb's Law |
 54%
From NCERT
PMT - 1996
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A positively charged ball hangs from a silk thread. We put a positive test charge q0 at a point and measure F/q0, then it can be predicted that the electric field strength E 

(1) > F/q0

(2) = F/q0

(3) < F/q0

(4) Cannot be estimated

Subtopic:  Electric Field |
From NCERT
PMT - 1990
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