Comment
Share
Q)

# A charge of 10^10C is placed at the origin the electric field at 1,1m due t

A charge of 10^10C is placed at the origin the electric field at 1,1m due to it is

Comment
A)

Q=10^-10

distance of electricf ield is (1,1)

so r=√(1^2+1^2)=√2

so E(r)=kQ/r

E=9*10^9*10^-10/√2

=0.636 N/C