JEE Main 9 Jan 2019 Shift 2 Solved Paper

Let

→

E1

&

→

E2

are the values of electric fields due to q1 & q2 respectively magnitude od E2 =

1

4π∊0

q2

r2

E2 =

9×109×(25)×10−6

(42+32)

V/m
E2 = 9 × 103 V/m
∴

→

E2

= 9 × 103 (cosθ2

^

i

−sinθ2

^

j

)
Since tan θ2 =

3

4

∴

→

E2

= 9 × 103(

4

5

^

i

−

3

5

^

j

) = (72

^

i

−54

^

j

) × 102
Magnitude of E1 =

1

4πε0

√10×10−6

(12+32)

= (9×109) × √10 × 10−7
= 9 √10×102
∴

→

E1

= 9 √10×102 [cosθ(−

^

i

)+sinθ1

^

j

]
∴ tan θ1 = 3

E1 = 9 × √10 × 102 [

1

√10

(−

^

i

)+

3

√10

^

j

]
E1 = 9 × 102[−

^

i

+3

^

j

] = [−9

^

i

+27

^

j

]102
∴

→

E

=

→

E1

+

→

E2

= (63

^

i

−27

^

j

)×102 V/m