Question

# Two point charges q and −q are at positions (0,0,d) and (0,0,−d) respectively. What will be the electric field at point (a,0,0) due to both the charges?

A
2kqd4π0(d2+a2)3/2^k
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B
qd4π0(d2+a2)3/2^k
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C
2kqd4π0(d2+a2)3/2^k
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D
kqd4π0(d2+a2)3/2^k
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Solution

## The correct option is C −2kqd4π∈0(d2+a2)3/2^k At P there are two fields, (1) Due to A, →EA=kqr2 Away from A (2) Due to B, →EB=kqr2 Towards B Distance of point P from both A and B, r=(a2+d2)1/2 Angle made by →Enet with →EA or →EB =12(180−2θ)=(90−θ) Net electric field at P, →Enet=2EAcos(90−θ) ⇒→Enet=2EAsinθ=2kqr2d(r) →Enet=2kqd(a2+d2)3/2 in −z direction →Enet=−2kqd(a2+d2)3/2^k

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