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}\$

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Electric Charges And Fields

Q: Two identical conducting spheres \$A\$ and \$B\$, carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is \$F\$. A third identical conducting sphere, \$C\$, is uncharged. Sphere \$C\$ is first touched to \$A\$, then to \$B\$, and the removed. As a result, the force between \$A\$ and \$B\$ would be equal to

Initially charges on both sphere, \$q\$ \$F = \dfrac{Kq^2}{d^2}\$ .............(1)When sphere C will get touched with sphere A, then final charges on both will become \$\dfrac{q}{2}\$.Now, when this sphere C will get touched to sphere B, then final charges on both of them will be \$q_c = q_d = \dfrac{q/2 + q}{2} = \dfrac{3q}{4}\$Now force between A and B will be \$F' = \dfrac{k \times \dfrac{q}{2} \times \dfrac{3q}{4}}{d^2}\$ \$F' = \dfrac{3F}{8}\$

Q: Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is:

Let initially B and C have charges \$Q\$ each and distance between them be \$r\$.\$F= \dfrac{Q^2}{4\pi \epsilon_o r^2}\$ Now when another uncharged spherical conductor A of same radius is brought in contact with B, then they will share equal charge i.e., A and B will have charge \$q=\dfrac{Q_1+Q_2}{2}\$ each.i.e \$q=\dfrac{Q}{2}\$Similarly, now when A is brought in contact with C, then both will have equal charge let say \$q'.\$\$q' = \dfrac{Q + \dfrac{Q}{2}}{2} = \dfrac{3Q}{4}\$Force between B and C, \$F' = \dfrac{\dfrac{Q}{2} \times \dfrac{3Q}{4}}{4\pi \epsilon_o r^2} = \dfrac{3F}{8}\$

Q: 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]\$

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 \$

Physics
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Two identical conducting spheres $A$ and $B$ , carry equal charge. They are separated by a distance much larger than their diameter, and the force between them is $F$ . A third identical conducting sphere, $C$ , is uncharged. Sphere $C$ is first touched to $A$ , then to $B$ , and the removed. As a result, the force between $A$ and $B$ would be equal to

\frac{3 F}{4}

\frac{F}{2}

F

\frac{3 F}{8}

Medium

JEE Mains

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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:

4 E h R

\frac{1}{2}

E h R

2 E h R

E h R

Medium

JEE Mains

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Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is:

\frac{F}{4}

\frac{3 F}{4}

\frac{F}{8}

\frac{3 F}{8}

Medium

JEE Mains

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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 $ϵ_{0} = 8.85 \times 1 0^{- 12} C^{2} / N - m^{2}, R_{E} = 6.4 \times 1 0^{6} m]$

+ 660 k C

- 660 k C

- 680 k C

+ 680 k C

Medium

JEE Mains

View solution>

Electric Charges And Fields

Physics 27672 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
27672 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: