AP EAPCET 24-Aug-2021 Shift 1 Paper

Given that, the point charges are placed on vertices of a square as shown in figure. Let r be the distance of O from the each comer.
In Δ‌ABC,AC2=AB2+BC2
(2r)2=a2+a2=2a2
⇒r=

a

√2

r=

2.8

√2

m (∵a=2.8m)
Now, using expression of electric potential at a point due to point charge
V=

Kq

r

Now, total electric potential at O due to all the four charges q1,q2q3 and q4
V0=V1+V2+V3+V4
V0=

Kq1

r

+

Kq2

r

+

Kq1

r

+

Kq4

r

V0=

K

r

[q1+q2+q3+q4]
Substituting the values, we get
Ve=

9×109×√2

(2.8)

(20+40−34+16)×10−9
=190.89V