Q: A cone of base radius \$R\$ and height \$h\$ is located in a uniform electric field \$\overrightarrow{E}\$ parallel to its base. The electric flux entering the cone is:

We know that,\$\displaystyle \phi = E.A Cos \theta\$Flux crossing through the area of cone parallel to the electric field which is nothing but a right angle triangle having base \$2R\$ and height \$h\$therefore, \$\displaystyle \phi= {EhR}\$

Question 1

A cone of base radius $R$ and height $h$ is located in a uniform electric field $\overrightarrow{E}$ parallel to its base. The electric flux entering the cone is:

Accepted Answer

We know that,$\displaystyle \phi = E.A Cos \theta$Flux crossing through the area of cone parallel to the electric field which is nothing but a right angle triangle having base $2R$ and height $h$therefore, $\displaystyle \phi= {EhR}$

Question 2

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 : [Given $\epsilon_0=8.85\times 10^{-12} C^2/N-m^2, R_E=6.4\times 10^6m]$

Accepted Answer

From Gauss' Law,the electric field entering the Earth's atmosphere is E. The flux of this field is,$ \int E.ds $ = $ E \times 4 \pi {R}^{2} $This should be equal to the charge enclosed inside the Earth divided by $ \epsilon $Hence,$ E \times 4 \pi {R}^{2} $ = $ \dfrac{q}{\epsilon} $Hence, putting back the value of $3$ knowns, the value of $q = -680 \times {10}^{3} C $

Electric Charges And Fields

Physics
4914 Views

A cone of base radius $R$ and height $h$ is located in a uniform electric field $E$ parallel to its base. The electric flux entering the cone is:

Electric Charges And Fields

Physics 4914 Views

A cone of base radius R and height h is located in a uniform electric field E parallel to its base. The electric flux entering the cone is:

Physics
4914 Views

A cone of base radius $R$ and height $h$ is located in a uniform electric field $E$ parallel to its base. The electric flux entering the cone is: