AP EAPCET 20-Aug-2021 Shift 2 Paper

Given, charges at point A and B are qA,qB=10µC and −10µC
Separation between A and B=10‌cm =10×10−2mLet O is mid-point of AB .
∴OA=5×10−2m⇒OP=12×10−2m
and AP=√122+52 =√169=13×10−2m
As we know that,
Electric field (E)=

kq

r2

where, k is Coulomb's constant =9×109C2m−2N−1 ,
q is charge and r is separation between A and P .
∴ Electric field due to charge A
EA=

9×109×10×10−6

(13×10−2)2

=5.3×106N∕C
Similarly, electric field due to charge at B,
EB=5.3×106N∕C
∴ Net Electric field (Enet)=2E‌sin‌θ =2×5.3×106×

5

13

=

53

13

×106
=4.07×106N∕C=4.1×106N∕C