CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
429
You visited us 429 times! Enjoying our articles? Unlock Full Access!
Question

In a regular polygon of n sides, each corner is at a distance r from the centre. Identical charges are placed at (n−1) corners. At the centre, the intensity is E and the potential is V. The ratio V/E has magnitude.

A
rn
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
r(n1)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
r
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
r(1)/n
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B r(n1)


Step 1: Electric potential

The electric potential is a scalar quantity

So, the potential at the center is the sum of potential due to all (n1) charges

Potential due to one charge q =kqr

So, Potential for (n1) charges. i.e, V=k(n1)qr ....(1)

Step 2: Electric field

The electric field is a vector quantity.

So the electric field cancels each other for the charges of the opposite corners of the polygon. (Refer figure)

For example, suppose n=8, there are 3 pairs of equal charges placed at corners so, the electric field of these pairs will be zero (equal and opposite)

Therefore only one charge q will contribute to the electric field at the center of the polygon.

Therefore, electric field E=kqr2 ....(2)

Step 3: Ratio of electric potential and electric field

From equation (1) and (2)

VE=k(n1)qr×r2kq=r(n1)

Hence, option B is correct


2109648_125080_ans_8d3c3d06f80442cb97f777c40737ccdb.png

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Applications of Gauss' Law and the Idea of Symmetry
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon