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%
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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 |
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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 |
 75%
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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%
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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%
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\(q_1, q_2,q_3~\text{and}~q_4\) are point charges located at points as shown in the figure and \(S\) is a spherical Gaussian surface of radius \(R\). Which of the following is true according to the Gauss’s law?


1. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3\right) \cdot d \vec{A}=\frac{q_1+q_2+q_3}{2 \varepsilon_0}\)
2. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3+\vec{E}_4\right) \cdot d \vec{A}=\frac{\left(q_1+q_2+q_3\right)}{\varepsilon_0}\)
3. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3\right) \cdot d \vec{A}=\frac{\left(q_1+q_2+q_3+q_4\right)}{\varepsilon_0}\)
4. \(\oint_s\left(\vec{E}_1+\vec{E}_2+\vec{E}_3+\vec{E}_4\right) \cdot d \vec{A}=\frac{\left(q_1+q_2+q_3+q_4\right)}{\varepsilon_0}\)

Subtopic:  Gauss's Law |
 77%
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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%
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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%
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