Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

Find the electric field potential and strength at the centre of a hemisphere of radius

R

charged uniformly with the surface density

σ

.

Accepted Answer

Given that radius of hemisphere is and surface density $σ .$

In order to find electric field strength and electric potential at the centre of hemisphere we have to take an elemental ring. Circumference of the elemental ring is located at an angle $θ$ from the horizontal plane and making angle $d θ$ at the centre.

Charge on of elemental ring $(d q) = σ \times Area of elemental ring$
$\Rightarrow d q = σ \{2 π r R d θ\} \Rightarrow d q = σ \{2 π (R \cos θ) R d θ\} \Rightarrow d q = 2 π σ R^{2} \cos θ d θ$

Due to this elemental ring, Potential $(d V)$ at the centre of hemisphere is given by,
$\Rightarrow d V = \frac{1}{4 π ε_{0}} \frac{d q}{R} \Rightarrow d V = \frac{1}{4 π ε_{0}} \frac{2 π σ R^{2} \cos θ d θ}{R} \Rightarrow d V = \frac{σ R}{2 ε_{0}} \cos θ d θ$
$∴$ Net electric Potential at the centre of hemisphere due to complete hemisphere is given by,

$\Rightarrow V_{net} = \int d V = \int_{0}^{\frac{π}{2}} \frac{σ R}{2 ε_{0}} \cos θ d θ \Rightarrow V_{net} = \frac{σ R}{2 ε_{0}} \int_{0}^{\frac{π}{2}} \cos θ d θ \Rightarrow V_{net} = \frac{σ R}{2 ε_{0}} {|\sin θ|}_{0}^{\frac{π}{2}} \Rightarrow V_{net} = \frac{σ R}{2 ε_{0}} \{\sin \frac{π}{2} - \sin 0\} \Rightarrow V_{net} = \frac{σ R}{2 ε_{0}}$

Due to this elemental ring, electric field $(d E)$ at the centre of hemisphere is given by,
$d E = \frac{1}{4 π ε_{0}} \frac{(d q) y}{{(r^{2} + y^{2})}^{\frac{3}{2}}} d E = \frac{1}{4 π ε_{0}} \frac{(2 π σ R^{2} \cos θ d θ) R \sin θ}{{\{{(R \cos θ)}^{2} + {(R \sin θ)}^{2}\}}^{\frac{3}{2}}} d E = \frac{1}{4 ε_{0}} \frac{\{σ R^{3} (2 \sin θ \cos θ) d θ\}}{R^{3}} d E = \frac{σ}{4 ε_{0}} \sin 2 θ d θ (Along - y - axis)$

$∴$ Net electric field strength at the centre of hemisphere is given by,
$E_{net} = \int d E = \int_{0}^{\frac{π}{2}} \frac{σ}{4 ε_{0}} \sin 2 θ d θ$
$E_{net} = \int d E = \int_{0}^{\frac{π}{2}} \frac{σ}{4 ε_{0}} \sin 2 θ d θ E_{_{net}} = \frac{σ}{4 ε_{0}} {|- \frac{\cos 2 θ}{2}|}_{0}^{\frac{π}{2}} E_{_{net}} = - \frac{σ}{8 ε_{0}} \{cosπ - \cos 0\} E_{net} = \frac{σ}{4 ε_{0}} (Along - y - axis)$

Find the electric field potential and strength at the centre of a hemisphere of radius $R$ charged uniformly with the surface density $σ$ .

Important Questions on Electrostatics

Learn and Prepare for any exam you want !

Find the electric field potential and strength at the centre of a hemisphere of radius R charged uniformly with the surface density σ.

Important Questions on Electrostatics

Learn and Prepare for any exam you want !

Find the electric field potential and strength at the centre of a hemisphere of radius $R$ charged uniformly with the surface density $σ$ .