Question

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

विकल्प

• 6.25 × 10⁷ N/C

• 9.28 × 10⁷ N/C

• 13.1 × 10¹¹ N/C

• 20.5 × 10⁷ N/C

Answer 1

6.25 × 10⁷ N/C

Explanation:

Given: Length of the dipole (2l) = 10 cm = 0.1 m 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)

= ${\displaystyle \frac{1}{4\pi {\epsilon }_0}\times \frac{2\left(\text{q}.2\text{l}\right)\text{r}}{{({\text{r}}^2-{\text{l}}^2)}^2}}$

= ${\displaystyle 9\times {10}^9\times \frac{2(500\times {10}^{-6}\times 0.1)\times 0.25}{{[{\left(0.25\right)}^2-{\left(0.05\right)}^2]}^2}}$

= 6.25 × 10⁷ N/C ...(k = 1 for air)