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:

पर्याय

• 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)

= `1/(4πε_0) xx (2("q".2"l")"r")/("r"^2 - "l"^2)^2`

= `9 xx 10^9 xx (2(500 xx 10^-6 xx 0.1) xx 0.25)/[(0.25)^2 - (0.05)^2]^2`

= 6.25 × 10⁷ N/C ...(k = 1 for air)