A point charge ‘q’ is placed at a distance ‘ℓ ’ (on$$ the $$y-axis) vertically above the semi infinite plane surface which lies in the xz plane as shown in the figure. The magnitude of flux of electric field passing through the surface is x $$q48$$ ∈$$0$$ where x is a positive integer. Find x ?

Question

A point charge ‘q’ is placed at a distance ‘ℓ ’ (on$$ the $$y-axis) vertically above the semi infinite plane surface which lies in the xz plane as shown in the figure. The magnitude of flux of electric field passing through the surface is x $$q48$$ ∈$$0$$  where x is a positive integer. Find x ?

Infinity Learn · Accepted Answer

ϕ$$=$$ϕ$$OAB+$$ϕABCD $$=q48$$ε$$0+q48$$ε$$0=q24$$ε$$0Imagine$$ a cuboid as shown. Flux through left and right face is zero, as electric filed at infinity is zero.ϕ$$cuboid=q$$ε$$0$$∴ϕ$$face=q4$$ε$$0$$⇒ϕ$$OABC=q16$$ε$$0Imagine$$ a cube as shown.ϕ$$cube=q$$ε$$0$$∴ϕ$$face=q6$$ε$$0$$⇒ϕ$$OAED=q24$$ε$$0$$⇒ϕ$$OAE=q48$$ε$$0Hence$$,ϕshaded $$region=$$ϕOABC−ϕ$$OAE=q16$$ε$$0$$−$$q48$$ε$$0=2q48$$ε$$0$$

Gauss's law and its applications

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Gauss's law and its applications

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A point charge ‘q’ is placed at a distance ‘ℓ ’ (on the y-axis) vertically above the semi infinite plane surface which lies in the xz plane as shown in the figure. The magnitude of flux of electric field passing through the surface is x q48 ∈0 where x is a positive integer. Find x ?

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free study material