Question

2. The electric intensity due to a dipole of length 10 cm and
having a charge of 500 uc, at a point on the axis at a
distance 20 cm from one of the charges in air, is
(a) 6.25 x 10'N/C (b) 9.28 x 10' N/C
(c) 13.1 x 11' N/C (d) 20.5 x 10'N/C​

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

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

$\huge\underline{\underline{\bf \orange{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Lenght of dipole (a) = 10cm or 0.1m
• Charge (q) = 500μC
• At point on axis at distance (l) = 20cm or 0.2m
• r = 0.20 + 0.5 = 0.25m

$\large\underline{\underline{\sf To\:Find:}}$

• Electric intensity (E)

$\huge\underline{\underline{\bf \orange{Solution-}}}$

✪$\large{\boxed{\bf \blue{E = \dfrac{1}{4π\epsilon_o}×\dfrac{2pr}{(r^2-l^2)^2}}}}$

$\large\implies{\bf p = q × a}$

$\implies{\sf p = 500×10^{-6}×0.1}$

$\implies{\sf \green{ p = 5 × 10^{-5} }}$

And , ${\sf \dfrac{1}{4π\epsilon_o}=9×10^9}$

$\implies{\sf E = 9×10^9×\dfrac{2×5×10^{-5}×0.25}{((0.25)^2-(0.5)^2)^2}}$

On solving this we get ⎯

$\implies{\bf \red{ E = 6.25×10^{7}N/C }}$

$\huge\underline{\underline{\bf \orange{Solution-}}}$

Electric intensity due to dipole is ${\bf \red{6.25×10^7\:N/C}}$ .

Answer 2

The electric intensity due to a dipole is 6.25 × 10⁷ N/C

Explanation:

The electric density is given by the formula:

E = 9 × 10⁹ . (2pr)/(r² - l²)²

Now, the value of the each variable is given as:

p = (500 × 10⁻⁶) × (10 × 10⁻²) = 5 × 10⁻⁵ Cm

r = 25 cm = 25 × 10⁻² m

l = 5 cm = 5 × 10⁻² m

On substituting value, we get,

E = 9 × 10⁹ . (2 × 5 × 10⁻⁵ × 25 × 10⁻²)/((25 × 10⁻²)² - (5 × 10⁻²)²)²

∴ E = 6.25 × 10⁷ N/C