Gauss's Law
Trending Questions
- Qϵ0
- 100Qϵ0
- 10Q(πϵ0)
- 100Q(πϵ0)
Why is the electric field uniform in parallel plates?
The electric flux for Gaussian surface A that enclose the charged particles in free space is (given q1=−14 nC, q2=78.85 nC, q3=−56 nC)
- 103 Nm2 c−1
- 103 CN−1 m−2
- 6.32×103 Nm2 C−1
- 6.32×103 CN−1 m−2
- q6ϵo
- q6ϵo
- q12ϵo
- q24ϵo
An electric dipole is placed at the center of an imaginary sphere. Which of the following option is/are correct?
The electric field is zero at every point of the sphere
The flux of the electric field at every point of the sphere is zero
The electric potential is zero at every point of the sphere
The electric potential is zero on a circle on the surface
(Take ϵo=9×10−12 C2/N-m2)
- 27.8 N/C
- 17.8 N/C
- 47.8 N/C
- 37.8 N/C
[Assume x to be very small as compared to dimension of sheet]
- Zero
- 10 N-m2/C
- 15 N-m2/C
- 5 N-m2/C
Is a cube a Gaussian Surface?
- Q3ε0
- 2Qε0
- Q2ε0
- Qε0
E=a(x^i+y^j)x2+y2
where, a is a constant and ^i and ^j are the unit vectors of the x and y− axes. Find the charge within a sphere of radius R with the centre at the origin.
A spherical volume contains a uniformly distributed charge of volume density 2.0×10−4 Cm−3. Find the strength of electric field at a point inside the volume at a distance 4.0 cm from the centre.
3×104 NC
3×105 NC
3×106 NC
3×107 NC
- ρr06ϵ0, towards left
- ρr06ϵ0, towards right
- 17ρr054ϵ0, towards right
- 17ρr054ϵ0, towards left
- ∮(→E1+→E2+→E3).d→A=q1+q2+q32ϵ0
- ∮(→E1+→E2+→E3).d→A=q1+q2+q3ϵ0
- ∮(→E1+→E2+→E3).d→A=q1+q2+q3+q4ϵ0
- None of the above
- 8eϵ0
- 16eϵ0
- eϵ0
- zero
- 5.65×109
- 1.13×1011
- 9×109
- 8.85×10−12
A charge q is located above the centre of a square plate of side a, at a distance a2. The flux of electric field through the plate Is?
qϵ0
q2ϵ0
q4ϵ0
q6ϵ0
E=a(x^i+y^j)x2+y2
where, a is a constant and ^i and ^j are the unit vectors of the x and y− axes. Find the charge within a sphere of radius R with the centre at the origin.
- −2πϵ0aR
- πϵ0aR
- 4πϵ0aR
- 2πϵ0aR
- 14
- 12
- 34
- 1
Comment on the correctness of the following two statements.
Statement I – Net Electric field strength on a Gaussian surface may change if the charges inside or outside move.
Statement II – Electric flux through a Gaussian surface will not change if the charges contained inside do not cross the boundary of the assumed Gaussian surface.
True, True
True, false
False, True
False, False
The electric flux through the surface S as due to all the charges as shown in following figure isq1+q3−q2−q4ϵ0
True
false
Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10 cm containing the total charge is 25 V-m. The flux through a concentric sphere of radius 20 cm will be?
25 V-m
50 V-m
100 V-m
200 V-m
- q1 and q2 are positive and q1<q2
- q1 and q2 are positive and q1>q2
- q1 and q2 are negative and |q1|<|q2|
- q1 is positive and q2 is negative ; q1<|q2|
- Zero.
- q14πε0r2 towards the center of cavity.
- q14πε0r21 away from the center of cavity.
- q14πε0r2 away from the center of cavity.
(1ϵo=4π×9×109)
- 162π×10−3Nm2C
- 162π×103Nm2C
- 162π×10−6Nm2C
- 162π×106Nm2C
- 14
- 12
- 34
- 1
Total electric flux coming out of a unit positive charge put in air is