Question

Two charge of $+25\times {10}^{-9}$ C and $-25\times {10}^{-9}$ C are placed 6 m apart. Find the electric field intensity ratio at point 4 m from the centre of the electric dipole

(i) on axial line

(ii) on equatorial line

Answer 1

Given that,

$q_1=+25\times {10}^{-9}C$

$q_2=-25\times {10}^{-9}C$

$2a=6m$

$a=3m$

(a) Electric field intensity at point 4 m from the centre on axial point is:

$E=\frac{1}{4\pi {\epsilon }_0}\frac{2pr}{{(r^2-a^2)}^2}$

$p=q\left(2a\right)$

$p=25\times {10}^{-9}\times 2\times 3$

$p=150\times {10}^{-9}$

$E=9\times {10}^9\frac{2\times 150\times {10}^{-9}\times 4}{{(4^2-3^2)}^2}$

$E=\frac{9\times {10}^9\times 2\times 150\times {10}^{-9}\times 4}{49}$

$E=\frac{10800}{49}N/C$

(b) Electric field intensity at point 4 m from the centre on equatorial point is:

$E=\frac{1}{4\pi {\epsilon }_0}\frac{p}{{(r^2+a^2)}^{3/2}}$

$E=\frac{9\times {10}^9\times 150\times {10}^{-9}}{{(4^2+3^2)}^{3/2}}$

$E=\frac{1350}{125}N/C$