E = ( x 2 + y 2 2 x ) i ^ + ( x 2 + y 2 2 y ) j ^
The electric field intensity in space varies according to above equation. Find the total charge enclosed in a sphere of radius 5 units which is centred at origin. If the answer is k coulombs, enter ⌊ 1 0 1 2 k ⌋ .
Details and assumptions:
i ^ and j ^ are unit vectors along x and y directions respectively.
Everything is in SI units.
Permittivity of free space, ϵ 0 = 8 . 8 5 4 × 1 0 − 1 2 .
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Problem treated in spherical coordinates:
x = R sin θ cos ϕ y = R sin θ sin ϕ z = R cos θ
r = x i ^ + y j ^ + z k ^ r ^ = ∣ r ∣ r
The electric field vector can be written in terms of the R , θ and ϕ . The surface area element on the sphere is:
d S = R 2 sin θ d θ d ϕ r ^
The elementary flux through the element is:
d Φ = E ⋅ d S
Plugging in expressions and simplifying gives: d Φ = 1 0 sin θ d θ d ϕ
The total flux through the sphere is:
Φ = ∫ 0 2 π ∫ 0 π 1 0 sin θ d θ d ϕ Φ = 4 0 π
According to Gauss' law:
Φ = ϵ o Q e n c
From here, the required answer can be computed.
⌊ 1 0 1 2 Q e n c ⌋ = 1 1 1 2
What I find strange about this problem is when I try to solve it using Maxwell's equation:
∇ ⋅ E = ϵ o ρ
Where ρ is the charge per unit volume in space. The divergence of the electric field evaluates to zero which implies that the charge per unit volume is zero, which implies that the total charge should be zero. I get a contradictory result using Maxwell's equation. Can someone clarify this for me? Thanks in advance
The only thing I can think of is that E is not differentiable at the origin.
Echoing Hosam, the Divergence Theorem only applies if the fields are continuously differentiable
Thank you for the clarification @Hosam Hajjir @Steven Chase
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At any point P ( x , y , z ) on the sphere, a unit vector perpendicular to the sphere radially outwards is n ^ = R x i ^ + R y j ^ + R z k ^ The electric flux passing through an elemental area d S at point P on the sphere can be given as d Φ = E ⋅ n ^ d S = ( 5 ( x 2 + y 2 ) 2 x 2 + 5 ( x 2 + y 2 ) 2 y 2 ) d S
Which implies d Φ = 5 2 d S . Since the flux is independent of coordinates, thus total flux can be integrated for whole sphere which is given as ∫ d Φ = Φ = 5 2 ∫ d S = 5 2 × 4 π ( 5 2 ) = 4 0 π
Hence from Gauss's law , charge enclosed can be given as 4 0 π ϵ 0 ≈ 1 . 1 1 2 × 1 0 − 9 .
Hence ⌊ 1 0 1 2 k ⌋ = 1 1 1 2 .