Two point charges q_1(√10 μ C) and q_2(-25 μ C) are placed on the x -axis at x=1 m and x=4 m respectively. The electric field ( in V/m) at a point…
JEE Main 2019 — Physics Electromagnetism
2019mcqhard
Two point charges q1(10μC) and q2(−25μC) are placed on the x -axis at x=1m and x=4m respectively. The electric field (inV/m) at a point y=3m on y-axis is,
[Take4πϵ01=9×109Nm2C−2]
Official previous-year question
Held on 9 Jan 2019 · Verified 6 Jul 2026.
Options
A
(−81i^+81j^)×102
B
(81i^−81j^)×102
C
(−63i^+27j^)×102
D
(63i^−27j^)×102
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Solution
Electric field due to 10μC
E1=4π∈01r1310×10−6r1
=4π∈01(10)310×10−6(−i^+3j^)
Similarly, electric field due to −25μC
E2=4π∈01(5)325×10−6(4i^−3j^)
Net electric field,
E=9×109×10−6(10−i^+103j^+54i^−53j^)
=9×103(107i^−103j^)
=(63i^−27j^)×102CN
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