Two point charges q1 (
) and q2(−25
are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,
a.
(−63
) × 102
b.
(81
-81
c.
(63
× 102
d.
(−81
×102
) and q2(−25 are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is, a.
(−63 ) × 102
b.
(81 -81
c.
(63 × 102
d.
(−81 ×102
Question ID - 50123 | SaraNextGen Top Answer Two point charges q1 ( ) and q2(−25 are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,
a.
(−63 ) × 102
b.
(81 -81
c.
(63 × 102
d.
(−81 ×102
Two point charges q1 ( ) and q2(−25 are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is,
|
a. |
(−63 |
b. |
(81 |
|
c. |
(63 |
d. |
(−81 |
1 Answer
127 votes
Answer Key / Explanation : (c) -
Let 
& 
are the vaues of electric field due to q1 and q2 Respectively magnitude of
E2 = 
E2= 
E2=9×103 V/m
∴ 
= 9×103 (
− sinθ2 
tanθ2 = 3/4
∴ 
= 9×103 
Magnitude of 
=
=(9
=9
102
∴ 
=9
102 [cos
1(
)+sin
1
]
∴ tan
=3
E1=9
102
E1=9
102
=
102
∴ 
=
1+
2 =(63
−27
)
102 V/m
∴correct answer is (c)
127 votes
Answer Link
127