Find the Charge!

E = ( 2 x x 2 + y 2 ) i ^ + ( 2 y x 2 + y 2 ) j ^ \vec{E}=\left(\dfrac{2x}{x^2+y^2}\right) \hat{i} + \left(\dfrac{2y}{x^2+y^2}\right) \hat{j}

The electric field intensity in space varies according to above equation. Find the total charge enclosed in a sphere of radius 5 5 units which is centred at origin. If the answer is k k coulombs, enter 1 0 12 k \lfloor 10^{12} k \rfloor .


Details and assumptions:

  • i ^ \hat{i} and j ^ \hat{j} are unit vectors along x x and y y directions respectively.

  • Everything is in SI units.

  • Permittivity of free space, ϵ 0 = 8.854 × 1 0 12 \epsilon_{0}=8.854 \times 10^{-12} .


The answer is 1112.

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2 solutions

Vilakshan Gupta
May 27, 2020

At any point P ( x , y , z ) P(x,y,z) on the sphere, a unit vector perpendicular to the sphere radially outwards is n ^ = x R i ^ + y R j ^ + z R k ^ \hat{n}=\dfrac{x}{R} \hat{i} + \dfrac{y}{R} \hat{j} + \dfrac{z}{R} \hat{k} The electric flux passing through an elemental area d S dS at point P P on the sphere can be given as d Φ = E n ^ d S = ( 2 x 2 5 ( x 2 + y 2 ) + 2 y 2 5 ( x 2 + y 2 ) ) d S d\Phi=\vec{E} \cdot \hat{n} dS= \left(\dfrac{2x^2}{5(x^2+y^2)}+\dfrac{2y^2}{5(x^2+y^2)}\right) dS

Which implies d Φ = 2 5 d S d\Phi=\dfrac{2}{5} dS . Since the flux is independent of coordinates, thus total flux can be integrated for whole sphere which is given as d Φ = Φ = 2 5 d S = 2 5 × 4 π ( 5 2 ) = 40 π \int d\Phi= \Phi=\dfrac{2}{5} \int dS=\dfrac{2}{5} \times 4\pi(5^2)=40\pi

Hence from Gauss's law , charge enclosed can be given as 40 π ϵ 0 1.112 × 1 0 9 40\pi\epsilon_{0} \approx 1.112 \times 10^{-9} .

Hence 1 0 12 k = 1112 \lfloor 10^{12}k \rfloor = \boxed{1112} .

Karan Chatrath
May 27, 2020

Problem treated in spherical coordinates:

x = R sin θ cos ϕ x = R \sin{\theta}\cos{\phi} y = R sin θ sin ϕ y = R \sin{\theta}\sin{\phi} z = R cos θ z = R\cos{\theta}

r = x i ^ + y j ^ + z k ^ \vec{r} = x \ \hat{i} + y \ \hat{j} +z \ \hat{k} r ^ = r r \hat{r} = \frac{\vec{r}}{\lvert \vec{r} \rvert}

The electric field vector can be written in terms of the R R , θ \theta and ϕ \phi . The surface area element on the sphere is:

d S = R 2 sin θ d θ d ϕ r ^ d\vec{S} = R^2\sin{\theta} \ d\theta \ d\phi \ \hat{r}

The elementary flux through the element is:

d Φ = E d S d\Phi = \vec{E} \cdot d\vec{S}

Plugging in expressions and simplifying gives: d Φ = 10 sin θ d θ d ϕ d\Phi = 10 \sin{\theta} \ d\theta \ d\phi

The total flux through the sphere is:

Φ = 0 2 π 0 π 10 sin θ d θ d ϕ \Phi = \int_{0}^{2 \pi} \int_{0}^{\pi} 10 \sin{\theta} \ d\theta \ d\phi Φ = 40 π \Phi = 40 \pi

According to Gauss' law:

Φ = Q e n c ϵ o \Phi = \frac{Q_{enc}}{\epsilon_o}

From here, the required answer can be computed.

1 0 12 Q e n c = 1112 \boxed{\lfloor 10^{12}Q_{enc} \rfloor= 1112}

What I find strange about this problem is when I try to solve it using Maxwell's equation:

E = ρ ϵ o \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_o}

Where ρ \rho 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.

Hosam Hajjir - 1 year ago

Echoing Hosam, the Divergence Theorem only applies if the fields are continuously differentiable

Steven Chase - 1 year ago

Thank you for the clarification @Hosam Hajjir @Steven Chase

Karan Chatrath - 1 year ago

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