Question

A point charge +Q is placed just outside an imaginary hemispherical surface of radius R . Find the electric flux passing through the curved surface of the hemisphere?

Answer 1

This is so interesting question. For solving this , you have to know about Gauss law, integration concepts.

Let's try to solve it .
Cut an element of radius r from of thickness dr as shown in figure.
E is the electric field along tangent of elementary ring
So, E = KQ/(√{R² + r²})^²
E = KQ/(R² + r²)
E = Q/4π∈₀(R² + r²)

Now, electric flux, Ф = E.ds
Here, dS = cosθ.dA
dA = 2πr.dr
And cosθ = R/√(R ² + r²)
∴ Ф = Q/4π∈₀(R² + r²). R/√(R² + r²) 2πr.dr
= QR/2∈₀ ∫ dr/(R² + r²)^(3/2)

Or, $\bold{\Phi=\frac{QR}{2\in_0}\int\limits^{R}_{0}\,\frac{r}{(R^2+r^2)^{3/2}}\,dr}$
$\bold{\Phi=\frac{QR}{2\in_0}[\large\frac{-1}{\sqrt{R^2+r^2}}]^R_0}$
= QR/2∈₀[ 1/R - 1/√2R ]
= Q/2∈₀ [ 1 - 1/√2]

Hence , electric flux passing through the curved surface of the hemisphere is Ф= Q/2∈₀ [ 1 - 1/√2]

Answer 2

Answer:

Explanation:

Its the easiest way