wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A point charge 50μC is located in the XY plane at the point of position vector r0=2^i+3^j. What is the electric field at the point vector r1=8^i5^j

A
1200V/m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.04V/m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
900V/m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4500V/m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 4500V/m

Step 1: Distance of charge from the point at which we have to find electric field [Refer Fig. 1]
Position vector of Charge Q, r0=2^i+3^j,
Position vector of Point where Electric Field has to be calculated, r1=8^i5^j

So, Distance between Point and Charge
r=|r|=|r1r0|=|8^i5^j2^i3^j|
=|6^i8^j|
r=62+82=10m

Step 2: Magnitude of Electric field [Refer Fig. 2]
Q=50 μC
By Coulomb's Law, Electric field of a point charge at a distance r
E=kQr2

=9×109×(50×106C)(10m)2=4500 V/m

Hence Correct Option id D

2111604_125175_ans_5d0518a6fb5340f58ed865874c274373.png

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Electron Sea Model_tackle
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon