Two point-charges ± 10μ C are placed 5.00 mm apart, forming an electric dipole. Compute electric field at a point on the axis of the dipole 15 cm away from the centre on a line passing through the centre dipole.

Question

Two point-charges ± 10μ C are placed 5.00 mm apart, forming an electric dipole. Compute electric field at a point on the axis of the dipole 15 cm away from the centre on a line passing through the centre dipole. (A)  1.66× 105NC-1  (B)  3.66× 105NC-1  (C)  2.66× 10-5NC-1  (D)  2.66× 105NC-1

Tardigrade Admin · Accepted Answer

The magnitude of the electric dipole moment is

p = q \times 2 l = (10 \times 1 0^{- 6}) \times (5.00 \times 1 0^{- 3})

= 50 \times 1 0^{- 9} c m

The field is required at point far-away from the centre of the dipole.
The electric field at an axial point (end-on position) distant

r (= 15 c m)

from the centre of the dipole is

E = \frac{1}{4 π ε}_{0} \frac{2 p}{r ^{3}} = (9.0 \times 1 0^{9}) \frac{2 \times ( 50 \times 1 0 ^{- 9} )}{( 15 \times 1 0 ^{- 2} ) ^{3}}

2.66 \times 1 0^{5} N C^{- 1}

Q. Two point-charges $\pm 10 μ C$ are placed $5.00 mm$ apart, forming an electric dipole. Compute electric field at a point on the axis of the dipole $15 c m$ away from the centre on a line passing through the centre dipole.

$1.66 \times 10^{5} N C^{- 1}$

$3.66 \times 10^{5} N C^{- 1}$

$2.66 \times 10^{- 5} N C^{- 1}$

$2.66 \times 10^{5} N C^{- 1}$

Q. Two point-charges ±10μC are placed 5.00mm apart, forming an electric dipole. Compute electric field at a point on the axis of the dipole 15cm away from the centre on a line passing through the centre dipole.

1.66×105NC−1

3.66×105NC−1

2.66×10−5NC−1

2.66×105NC−1

Q. Two point-charges $\pm 10 μ C$ are placed $5.00 mm$ apart, forming an electric dipole. Compute electric field at a point on the axis of the dipole $15 c m$ away from the centre on a line passing through the centre dipole.

$1.66 \times 10^{5} N C^{- 1}$

$3.66 \times 10^{5} N C^{- 1}$

$2.66 \times 10^{- 5} N C^{- 1}$

$2.66 \times 10^{5} N C^{- 1}$