Question

4. A very long wire contains a uniform negative charge density ( -lambda ). A charge ( -q ) rests inside a spherical Gaussian surface of radius R. The sphere's center lies on the wire. Located outside the sphere is another charge ( +q ). What is the total electric flux through the sphere? (5 points) (a) ( Phi=0 ) (b) ( Phi=-q / varepsilon_{0} ) (c) ( Phi=-k q / R^{2} ) (d) ( Phi=(-2 lambda R-q) / varepsilon_{0} ) (e) ( Phi=mathrm{k}(lambda+q) / varepsilon_{0} )

          4. A very long wire contains a uniform negative charge density ( -lambda ). A charge ( -q ) rests inside a spherical Gaussian surface of radius R. The sphere's center lies on the wire. Located outside the sphere is another charge ( +q ). What is the total electric flux through the sphere? (5 points)
(a) ( Phi=0 )
(b) ( Phi=-q / varepsilon_{0} )
(c) ( Phi=-k q / R^{2} )
(d) ( Phi=(-2 lambda R-q) / varepsilon_{0} )
(e) ( Phi=mathrm{k}(lambda+q) / varepsilon_{0} )

4. A very long wire contains a uniform negative charge density ( -lambda ). A charge ( -q ) rests inside a spherical Gaussian surface of radius R. The sphere's center lies on the wire. Located outside the sphere is another charge ( +q ). What is the total electric flux through the sphere? (5 points)
(a) ( Phi=0 )
(b) ( Phi=-q / varepsilon0 )
(c) ( Phi=-k q / R^2 )
(d) ( Phi=(-2 lambda R-q) / varepsilon0 )
(e) ( Phi=mathrmk(lambda+q) / varepsilon0 )

Added by Ceren D.

Master Resource Book in JEE Main Physics

D. B. Singh 1st Edition

Chapter 17

Instant Answer

Solved by Expert Ankur S

06/19/2023

Step 1

According to Gauss's Law, the electric flux through a closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space (ε₀). Mathematically, this is represented as Φ = Q_enclosed / ε₀. Show more…

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