Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Two infinite planes each with uniform surface charge density + are kept in each such a way that the angle between them is 300. The electric field in the region shown

Two infinite planes each with uniform surface charge density +\sigma are kept in each such a way that the angle between them is 300. The electric field in the region shown between them is given by:
Option: 1 \frac{\sigma }{\epsilon _{0}}\left [ \left ( 1+\frac{\sqrt{3}}{2} \right )\widehat{y}+\frac{\widehat{x}}{2} \right ]
   
   
Option: 2 \frac{\sigma }{2\epsilon _{0}}\left [ \left ( 1+{\sqrt{3}} \right )\widehat{y}+\frac{\widehat{x}}{2} \right ]

Option: 3 \frac{\sigma }{2\epsilon _{0}}\left [ \left ( 1+{\sqrt{3}} \right )\widehat{y}-\frac{\widehat{x}}{2} \right ]  
  
Option: 4 \frac{\sigma }{2\epsilon _{0}}\left [ \left ( 1-\frac{\sqrt{3}}{2} \right )\widehat{y}-\frac{\widehat{x}}{2} \right ]
 

Answers (1)

best_answer

 

 

Electric field due to continuous charge distribution -

So, let us consider a rod of length l which has uniformly positive charge per unit length lying on x-axis, dx is the length of one small section. This rod is having a total charge Q and dq is the charge on dx segment. The charge per unit length of the rod is \lambda. We have to calculate the electric field at a point P which is located along the axis of the rod at a distance of 'a' from the nearest end of Rod as shown in the figure - 

 

The field  \vec{dE} at P due to each segment of charge on the rod is in the negative "x" direction because every segment of the rod carry positive charge. In this every segment of the rod is producing electric field in the negative "x" direction, so the sum of electric field can be added directly and can be integrated because all electric field lies in the same direction. 

Now, here - dq = \lambda.dx

 

\begin{array}{l}{\qquad d E=k_{\mathrm{e}} \frac{d q}{x^{2}}=k_{\mathrm{e}} \frac{\lambda d x}{x^{2}}} \\ \\{\text { The total lield at } P \text { is }} \\ \\ {\qquad E=\int_{a}^{l+a} k_{e} \lambda \frac{d x}{x^{2}}} \\ \\ {\text { If } k_{e} \text { and } \lambda=Q/l \text { are constants and can be removed from the }} {\text { integral, then }} \\ \\ {\qquad \begin{aligned} E &=k_{e} \lambda \int_{a}^{l+a} \frac{d x}{x^{2}}=k_{e} \lambda\left[-\frac{1}{x}\right]_{a}^{l+a} \\ \\ \Rightarrow k_{e} \frac{Q}{l}\left(\frac{1}{a}-\frac{1}{l+a}\right)=\frac{k_{e} Q}{a(l+a)} \end{aligned}}\end{array}

Now if we slide the rod toward the origin and the a\rightarrow 0, then due to that end, the electric field is infinite.  

-

 

 

Assuming sheet to be non conducting

E_{net}=E((1-\frac{\sqrt{3}}{2})\hat j-\frac{1}{2}\hat i)

So,

E=\frac{\sigma}{2\epsilon_0}

So,

E_{net}=\frac{\sigma}{2\epsilon_0}((1-\frac{\sqrt{3}}{2})\hat y-\frac{1}{2}\hat x)

Where y and x are corresponding unit vectors

 

So option (4) is correct.

Posted by

Ritika Jonwal

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 registration ends tomorrow; changes introduced in exam this year
anam-khan

BTech Admission 2025 Live: JEE Mains registrations closing tomorrow; VITEEE, MET exam dates out
anam-khan

JEE Main 2025 Live: NTA to close JEE registration on Friday; apply at jeemain.nta.nic.in, correction window

Latest Question