A line of charge extends along a X-axis whose linear charge density varies directly as x. Imagine a spherical volume with its centre located on the X-axis and is moving gradually along it. Which of the graphs shown in figure correspond to the flux with the x coordinate of the centre of the volume?
(A). a
(B). b
(C). c
(D). d
Answer
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Hint: The linear density of charge (λ) is the amount of charge per unit length measured in coulombs per meter, on a line charge distribution at any point. The density of charges can be either positive or negative, as electric charges can be positive or negative. Flux describes any effect that a surface or substance seems to pass or travel through. In applied mathematics and vector calculus, a flux is a motion. Calculate linear charge density and then calculate Flux of the sphere.
Formula used:
$\phi =\dfrac{\lambda }{2\pi {{\varepsilon }_{0}}\text{r}}\cdot 4\pi {{\text{r}}^{2}}$
Complete step by step solution:
The total flow through the sphere, being a scalar quantity, will be equal to the algebraic sum of all these flows, i.e. This expression demonstrates that \[\dfrac{1}{{{e}_{0}}}\] times the charge enclosed (q) in the sphere is the total flux through the sphere.
Given that linear charge density $\lambda$ varies directly to $x$
$\Rightarrow \lambda \alpha x$
$=\lambda=\mathrm{kx}$
$\mathrm{k}=$ proportionality constant electric field due to long straight wire $r=$ distance of point $\mathrm{E}=\dfrac{\lambda}{2 \pi \varepsilon_{0} \mathrm{r}}$
initially when sphere with its center at $x$ -axis has flux $=\varepsilon$.A
$\phi=\dfrac{\lambda}{2 \pi \varepsilon_{0} \mathrm{r}} \cdot 4 \pi \mathrm{r}^{2}=2 \lambda \mathrm{r}$
as radius is constant
$\Rightarrow \phi \propto \lambda \text{ }$
$\Rightarrow \phi \propto x$
\[\therefore \] Flux is $\propto x$ it’s a straight-line motion
Hence, the correct option is (C).
Note:
Flux is a vector quantity for transport phenomena, describing the magnitude and the direction of the flow of a substance or property. Flux is a scalar quantity in vector calculus, defined as the surface integral of the perpendicular component of a vector field over a surface. The electric field immediately above a conductor's surface is directed to that surface normally. Now, as all the charge lies on the shell, the Gaussian surface includes no charge, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero.
Formula used:
$\phi =\dfrac{\lambda }{2\pi {{\varepsilon }_{0}}\text{r}}\cdot 4\pi {{\text{r}}^{2}}$
Complete step by step solution:
The total flow through the sphere, being a scalar quantity, will be equal to the algebraic sum of all these flows, i.e. This expression demonstrates that \[\dfrac{1}{{{e}_{0}}}\] times the charge enclosed (q) in the sphere is the total flux through the sphere.
Given that linear charge density $\lambda$ varies directly to $x$
$\Rightarrow \lambda \alpha x$
$=\lambda=\mathrm{kx}$
$\mathrm{k}=$ proportionality constant electric field due to long straight wire $r=$ distance of point $\mathrm{E}=\dfrac{\lambda}{2 \pi \varepsilon_{0} \mathrm{r}}$
initially when sphere with its center at $x$ -axis has flux $=\varepsilon$.A
$\phi=\dfrac{\lambda}{2 \pi \varepsilon_{0} \mathrm{r}} \cdot 4 \pi \mathrm{r}^{2}=2 \lambda \mathrm{r}$
as radius is constant
$\Rightarrow \phi \propto \lambda \text{ }$
$\Rightarrow \phi \propto x$
\[\therefore \] Flux is $\propto x$ it’s a straight-line motion
Hence, the correct option is (C).
Note:
Flux is a vector quantity for transport phenomena, describing the magnitude and the direction of the flow of a substance or property. Flux is a scalar quantity in vector calculus, defined as the surface integral of the perpendicular component of a vector field over a surface. The electric field immediately above a conductor's surface is directed to that surface normally. Now, as all the charge lies on the shell, the Gaussian surface includes no charge, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero.
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