Gauss’s law is true only if force due to a charge varies as
Gauss’s law is true only if force due to a charge varies as
- A
${r^{ - 1}}$
- B
${r^{ - 2}}$
- C
${r^{ - 3}}$
- D
${r^{ - 4}}$
Similar Questions
A square surface of side $L$ metres is in the plane of the paper. A uniform electric field $\vec E(V/m) $, also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is
A square surface of side $L$ metres is in the plane of the paper. A uniform electric field $\vec E(V/m) $, also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is
- [AIPMT 2006]
Gauss’s law should be invalid if
Gauss’s law should be invalid if
The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $150\, N/C$, directed inward towards the center of the Earth . This gives the total net surface charge carried by the Earth to be......$kC$ [Given ${\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}/N - {m^2},{R_E} = 6.37 \times {10^6}\,m$]
The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $150\, N/C$, directed inward towards the center of the Earth . This gives the total net surface charge carried by the Earth to be......$kC$ [Given ${\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}/N - {m^2},{R_E} = 6.37 \times {10^6}\,m$]
- [JEE MAIN 2014]
A rectangular surface of sides $10 \,cm$ and $15 \,cm$ is placed inside acyniform electric field of $25 \,V / m$, such that the surface makes an angle of $30^{\circ}$ with the direction of electric field. Find the flux of the electric field through the rectangular surface .................. $Nm ^2 / C$
A rectangular surface of sides $10 \,cm$ and $15 \,cm$ is placed inside acyniform electric field of $25 \,V / m$, such that the surface makes an angle of $30^{\circ}$ with the direction of electric field. Find the flux of the electric field through the rectangular surface .................. $Nm ^2 / C$
A charge $q$ is surrounded by a closed surface consisting of an inverted cone of height $h$ and base radius $R$, and a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{n q}{6 \epsilon_0}$ (in SI units). The value of $n$ is. . . .
A charge $q$ is surrounded by a closed surface consisting of an inverted cone of height $h$ and base radius $R$, and a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{n q}{6 \epsilon_0}$ (in SI units). The value of $n$ is. . . .
- [IIT 2022]