Question:

The electric field intensity at a point on the axis of an electric dipole in ah is 4  NC14 \; NC^{-1}. Then the electric field intensity at a point on the equitorial hue which is at a distance equal to twice the distance on the axial line and if the dipole is in a medium of dielectric constant 44 is

Updated On: Apr 4, 2024
  • 1  NC11 \; NC^{-1}
  • 18  NC1\frac{1}{8} \; NC^{-1}
  • 16    NC1 16 \; \; NC^{-1}
  • 116  NC1\frac{1}{16} \; NC^{-1}
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The Correct Option is D

Solution and Explanation

As, electric field intensity on the axis,
Eaxis =2kPr3E_{\text {axis }}=\frac{2 k P}{r^{3}}
4=2kPr3 \Rightarrow 4=\frac{2 k P}{r^{3}} for r>>a r > >a
( given, Eaxis =4) \left(\because \text { given, } E_{\text {axis }}=4\right)
kPr3=2\Rightarrow \frac{k P}{r^{3}}=2 \ldots (i)
Electric field intensity on equatorial line,
Eeq=kPr13E_{ eq }=\frac{k^{\prime} P}{r_{1}^{3}}
where, r1=2rr_{1}=2 r and k=k4 k'=\frac{k}{4}
So, Eeq=kP4×8r3E_{ eq }=\frac{k P}{4 \times 8 r^{3}}
Now, from the E (i), we get
Eeq=24×8=116NC1E_{ eq }=\frac{2}{4 \times 8}=\frac{1}{16} NC ^{-1}
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Concepts Used:

Electric Dipole

An electric dipole is a pair of equal and opposite point charges -q and q, separated by a distance of 2a. The direction from q to -q is said to be the direction in space.

p=q×2a

where,

p denotes the electric dipole moment, pointing from the negative charge to the positive charge.

Force Applied on Electric Dipole