Problem in electrostatics: E-field near 2 point charges

  • #1

Homework Statement


Refer the image.

Homework Equations


kq1q2/r^2 = F
Potential energy = kq1q1/r

The Attempt at a Solution


Obviously since both charges are unequal in magnitude option a is incorrect.
Calculating field at large distance r,
E = kq1/r^2 - kq2/r^2
= kq2/r^2
Also potential energy at q1 is lesser than near q2 for negative charge.
So options c,d are correct. However I don't understand how the angle beta can be calculated from the given information.
 

Attachments

  • Screenshot-2018-5-26 master_A dvi - NEST2016_A pdf.png
    Screenshot-2018-5-26 master_A dvi - NEST2016_A pdf.png
    39.2 KB · Views: 335

Answers and Replies

  • #2
However I don't understand how the angle beta can be calculated from the given information.
No need to calculate the angle to disqualify answer b.
 
  • #3
No need to calculate the angle to disqualify answer b.
I'm sorry but I don't quite understand. The answers given are b,c,d.
 
  • #4
Hmm. What book is this from?
 
  • #5
Hmm. What book is this from?
It's from an exam paper called NEST (National Entrance Screening Test) conducted in India. Its for the year 2016.
 

Suggested for: Problem in electrostatics: E-field near 2 point charges

Replies
3
Views
355
Replies
4
Views
626
Replies
3
Views
395
Replies
6
Views
405
Replies
16
Views
385
Replies
2
Views
460
Back
Top