Name: Dielectrics in a Capacitor
Uploaded: 2019-07-19
Duration: 6 min 3 s
Description: Let us use our knowledge about dielectrics and capacitors to find out how the use of dielectrics can increase the capacitance of a system. Learn through this...

Question

A sphere of radius

r

carries a surface charge of density

σ = \vec{a} \cdot \vec{r}

, where

\vec{a}

is a constant vector and

\vec{r}

is the radius vector of a point on the surface of the sphere, relative to its centre. Find the electric field strength vector at the centre of the sphere.

Accepted Answer

In the problem, the surface charge density of the sphere is given as $σ = \vec{a} \cdot \vec{r}$ , where we have assumed the direction of the constant vector $\vec{a}$ towards right, as shown in the figure. For any point on the sphere, with an angular position $θ$ , the surface charge density is,

$σ = a r cosθ$

Let us consider a very thin circular strip (ring) on the surface of the sphere. The half-angle subtended by this strip at the centre of the sphere is $θ$ . It is worth noting that the surface charge density at all the points on this strip is the same. The amount of charge on the strip will be,

$d q = σ (2 π r sinθ) r dθ$

$d q = 2 π a r^{3} sinθ cosθ dθ$

The strength of electric field at the centre, due to the strip/ring, will be

$d E = \frac{k (d q) x}{{(R^{2} + x^{2})}^{\frac{3}{2}}} = \frac{d q r cosθ}{4 π ε_{0} {(r^{2} \sin^{2} θ + r^{2} \cos^{2} θ)}^{\frac{3}{2}}}$

$d E = \frac{a r}{2 ε_{0}} sinθ \cos^{2} θ dθ$

Integrating the above expression within proper limits, we get,

$E = \frac{a r}{2 ε_{0}} \int_{0}^{π} sinθ \cos^{2} θ dθ$

$E = \frac{a r}{2 ε_{0}} (\frac{2}{3}) = \frac{a r}{3 ε_{0}}$

In order to find the direction of the electric field, we need to understand the nature of charge on different parts of the sphere. As we know that the cosine function is always positive in the first quadrant and always negative in the second quadrant, so the right half of the sphere will be positively charged and the left half would be negatively charged. Hence, the direction of the electric field will be opposite to that of the vector $\vec{a}$ .

$\vec{E} = - \frac{\vec{a} r}{3 ε_{0}}$

A sphere of radius $r$ carries a surface charge of density $σ = \vec{a} \cdot \vec{r}$ , where $\vec{a}$ is a constant vector and $\vec{r}$ is the radius vector of a point on the surface of the sphere, relative to its centre. Find the electric field strength vector at the centre of the sphere.

Important Questions on ELECTRODYNAMICS

Learn and Prepare for any exam you want !

A sphere of radius r carries a surface charge of density σ=a→·r→, where a→ is a constant vector andr→ is the radius vector of a point on the surface of the sphere, relative to its centre. Find the electric field strength vector at the centre of the sphere.

Important Questions on ELECTRODYNAMICS

Learn and Prepare for any exam you want !

A sphere of radius $r$ carries a surface charge of density $σ = \vec{a} \cdot \vec{r}$ , where $\vec{a}$ is a constant vector and $\vec{r}$ is the radius vector of a point on the surface of the sphere, relative to its centre. Find the electric field strength vector at the centre of the sphere.