Question

calculate the electric field at 60 cm from the centre of dipole on the axial line and the equatorial line​

Answer 1

The method and answer is in the picture.

Answer 2

Given ,

The two charges 10 uc and -10 uc are separated by distance 2 m

We know that , the dipole moment is the cross product of either charge (q) and dipole length (2a)

$\mathtt{\large \fbox{P = q \times 2a}}$

Thus ,

P = 10 × (10)^-6 × 2 × (10)^-2

P = 2 × (10)^-7

P = 2 × (10)^-7 C•m

We know that , the dipole field at an equitorial point at distance r from the centre of dipole is given by

$\large \mathtt{\fbox{E = k\frac{p}{ {(r)}^{3} } }}$

Thus ,

$\sf \hookrightarrow E = \frac{9 \times {(10)}^{9} \times 2 × {(10)}^{-7} }{ {(60 \times {(10)}^{ -2}) }^{3} } \\ \\\sf \hookrightarrow E = \frac{18 \times {(10)}^{2} }{216 \times {(10)}^{ - 2} } \\ \\ \sf \hookrightarrow E = \frac{18 \times {(10)}^{4} }{216} \\ \\\sf \hookrightarrow E = 0.083 \times {(10)}^{4} \: \: n/c$

We know that , at same distance

$\large \mathtt{\fbox{E_{(axial)} = 2 \times E_{(equitorial )}}}$

Thus ,

$\sf \hookrightarrow E = 2 × 0.083 × {(10)}^{4} \\ \\ \sf \hookrightarrow </p><p>E = 0.0166× {(10)}^{4} \\ \\ \sf \hookrightarrow </p><p>E = 166 \: \: n/c$