The electric field 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

A
6.25 x 10⁷ N/C
B
9.28 x 10⁷ N/C
C
13.1x11¹¹ N/C
D
20.5 x 10⁷ N/C

Solution:

$We know, E = 9 \times 10^{9} \cdot \frac{2 pr}{{(r^{2} - l^{2})}^{2}}$

$\begin{array}{rcl} where p & = & (500 \times 10^{- 6}) \times (10 \times 10^{- 2}) = 5 \times 10^{- 5} Cm \\ r & = & 25 cm = 0.25 m, l = 5 cm = 0.05 m \\ E & = & \frac{9 \times 10^{9} \times 2 \times 5 \times 10^{- 5} \times 0.25}{{\{(0.25)^{2} - (0.05)^{2}\}}^{2}} \\ = & 6.25 \times 10^{7} N / C \end{array}$

The electric field 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

Solution:

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