COMEDK Exam 2012 Solved Paper

Given,
Q1=2µC=2×10−6C
Q2=1µC=1×10−6
The electric field intensity at a point P due to a point charge is given by
E=

1

4πε0

Q

r2

⇒E∝

Q

r2

∴ Ratio of electric fields due to Q1 and Q2 at point P is

E1

E2

+

Q1

r12

×

r22

Q2

=

2×10−6

(2×10−2)2

×

(1×10−2)

1×10−6

=

1

2

Here, E1 is along X - axis and E2 is along Y - axis, So, the resultant field E have angle θ with X-axis as
tanθ=

E2

E1

=2