Question

The electric intensity due to a dipole of length 10 cm and having a charge of 500 mC, at a point on the axis at a distance 20 cm from one of the charges in air, is

(A) 6.25 x 10⁷ N/C
(B) 9.28 x 10⁷ N/C
(C) 13.1 x 10¹¹ N/C
D) 20.5x 10⁷ N/C

Hint - A is correct​

Answer 1

EXPLANATION.

Electric intensity due to a dipole of length = 10cm.

⇒ Charge = 500μc.

At the point on the axis distance = 20cm.

As we know that,

Formula of Dipole moment = 2ql.

Length of dipole = 2l = 10cm = 0.1m.

⇒ Length of dipole = l = 0.1/2 = 0.05m.

⇒ P = 2x 500 x 10⁻⁶ x 0.1.

⇒ P = 5 x 10⁻⁵.

Distance between center of pole and the point.

⇒ r = 20cm.

⇒ r = 20 + 5 = 25cm.

⇒ r = 0.25m.

As we know that,

⇒ E = 2kpr/(r² - l²)².

⇒ k = 1/4πεο = 9 x 10⁹ Nm²/C².

⇒ E = 2 x 9 x 10⁹ x 5 x 10⁻⁵ x 0.25/[(0.25)² - (0.05)²]².

⇒ E = 90 x 10⁴ x 0.25/[(0.0625 - 0.0025)]².

⇒ E = 90 x 10000 x 25/100/[0.06]².

⇒ E = 90 x 100 x 25 / 0.0036.

⇒ E = 225000/0.0036 = 6.25 x 10⁷ N/C.

Option [A] is correct answer.

Answer 2

$\mathfrak\red{Answer is 'a'}$

Length of the dipole (2l) =10cm = 0.1m or l = 0.05 m

Charge on the dipole (q) = 500 µC = 500 ×10–6 C and distance of the point on the axis from the mid-point of the dipole (r) = 20 + 5 = 25 cm = 0.25 m.

We know that the electric field intensity due to dipole on the given point

$(e) = \frac{1}{4\pi e_{o} } \times \frac{2(q.2)r}{( {r}^{2} - {l}^{2}) }$

$= 9. {10}^{9} \times \frac{2(500 \times {10}^{ - 60.001} )}{(0.25) ^{2} .(0.05) ^{2} ) ^{2} }$

$= \frac{225 \times {10}^{3} }{3.6 \times {10}^{3} }$

$= 6.25 \times {10}^{7} \frac{n}{s}$