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Electric Flux & Gauss Law

A 120 μ C poi...

Question

A 120 $μ C$ point charge is located at origin . The magnitude of electric flux passing through the portion of spherical shape defined by r = 10 cm, bounded by $θ = 0 t o π$ and $ϕ = 0 t o \frac{π}{2}$ rad will be _____ $μ C$ (upto 2 decimal place).

30

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Solution

The correct option is A 30
Electric flux density , $\to D = \frac{Q}{4 π r^{2}} {^a}_{r} C / m^{2}$

For electric flux crossing sphericla defined by

r= 10cm , $0 < θ \leq π 0 < ϕ \leq \frac{π}{2}$

$Ψ = \int \int \to D . \to d s$

$Ψ |_{r = 0.1 m} = \int \int \frac{Q}{4 π r^{2}} {^a}_{r} . r^{2} s i n θ d θ d ϕ_{r}$

= $\frac{Q}{4 π} \int_{θ = 0}^{π} s i n θ d θ \int_{ϕ = 0}^{\frac{π}{2}} d ϕ$

$= \frac{120 \times 10^{- 6}}{4 π} \times 2 \times \frac{π}{2}$
= 30.00 $μ C$

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A 120 μC point charge is located at origin . The magnitude of electric flux passing through the portion of spherical shape defined by r = 10 cm, bounded by θ=0 to π and ϕ=0 to π2 rad will be _____ μC (upto 2 decimal place). 30

The correct option is A 30Electric flux density , →D=Q4πr2^arC/m2 For electric flux crossing sphericla defined by r= 10cm , 0<θ≤π 0<ϕ≤π2 Ψ=∫∫→D.→ds Ψ|r=0.1m=∫∫Q4πr2^ar.r2sinθdθdϕ^ar = Q4π∫πθ=0sinθdθ∫π2ϕ=0dϕ =120×10−64π×2×π2 = 30.00 μC

A 120 $μ C$ point charge is located at origin . The magnitude of electric flux passing through the portion of spherical shape defined by r = 10 cm, bounded by $θ = 0 t o π$ and $ϕ = 0 t o \frac{π}{2}$ rad will be _____ $μ C$ (upto 2 decimal place).

30