First slide

Electric field

Question

A thin glass rod is bent into a semicircle of radius r. A charge +Q is uniformly distributed along the upper half, and a charge –Q is uniformly distributed along the lower half, as shown in the figure. The resultant electric field E at P, the center of the semicircle, is

Very difficult

A

$\frac{Q}{π^{2} ε_{0} r^{2}}$
B

$\frac{2 Q}{π^{2} ε_{0} r^{2}}$
C

$\frac{4 Q}{π^{2} ε_{0} r^{2}}$
D

$\frac{Q}{4 π^{2} ε_{0} r^{2}}$

Solution

The magnetic of the field at p due to either element is
$\begin{array}{l} d E = \frac{1}{4 π ε_{0}} \frac{r d θ \times Q / (π r / 2)}{r^{2}} \\ = \frac{Q}{2 π^{2} ε_{0} r^{2}} d θ \end{array} w h e r e θ i s t h e a n g l e m a d e b y t h e t w o r a d i u s v e c t o r o n b o t h s i d e s o f h o r i z o n t a l a n d r d θ a r e t h e t w o e l e m e n t s o n b o t h t h e s i d e s$
$∴$ field at P due to pair of elements is $2 d E \sin θ$
$i . e ., E = \int_{0}^{π / 2} 2 d E \sin θ$

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