Two point charges of 1 μ C and -1 μ C are separated by a distance of 100 mathringA. A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be

Question

Two point charges of 1 μ C and -1 μ C are separated by a distance of 100 mathringA. A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be (A) 9 N C-1 (B) 0.9 N C-1 (C) 90 N C-1 (D) 0.09 N C-1

Tardigrade Admin · Accepted Answer

The point lies on equatorial line of a short dipole. ∴ E=(2ql/4πε0r3) =(9×109×10-6×10-8/(10-1)3) =0.09 N C-1

Q. Two point charges of $1 μ C$ and $- 1 μ C$ are separated by a distance of $100 \overset{˚}{A}$ . A point $P$ is at a distance of $10 c m$ from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at $P$ will be

$9 N C^{- 1}$

$0.9 N C^{- 1}$

$90 N C^{- 1}$

$0.09 N C^{- 1}$

The point lies on equatorial line of a short dipole.
$∴ E = \frac{2 ql}{4 π ε _{0} r ^{3}}$
$= \frac{9 \times 1 0 ^{9} \times 1 0 ^{- 6} \times 1 0 ^{- 8}}{( 1 0 ^{- 1} ) ^{3}}$
$= 0.09 N C^{- 1}$

Q. Two point charges of 1μC and −1μC are separated by a distance of 100A˚. A point P is at a distance of 10cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be

9NC−1

0.9NC−1

90NC−1

0.09NC−1