Question

Two charges of +25 nC and -25 nC are placed 6 m apart.
Find the intensity of electric field at a point 4 m from the
centre of electric dipole
(i) on axial line
(ii) on equitorial line​

Answer 1

$\rm \huge{\fbox{Answer :}}$

(i) 4.2 × (10)^10 N/c

(ii) 2.1 × (10)^10 N/c

$\rm \huge{\fbox{Explaination :}}$

Given that ,

The two charges of +25 nC and -25 nC are placed 6 m apart

We know that ,

The dipole field at an equitorial at distance r from the centre of dipole is given by

$\sf \large \fbox{E = K\frac{p}{ {(r)}^{3} } }$

Thus ,

$\sf \mapsto E = 9 \times {(10)}^{9} \times \frac{25 \times 6}{ {(4)}^{2} } \\ \\ \sf \mapsto E = \frac{135 \times {(10)}^{10} }{64} \\ \\ \sf \mapsto E = 2.1 \times {(10)}^{10} \: \: N/c$

We know that ,

The dipole field at an axial point at distance r from the centre of dipole is given by

$\sf \large \fbox{E = 2( K\frac{p}{ {(r)}^{3} } )}$

Thus ,

$\sf \mapsto E = 2 \times 2.1 \times {(10)}^{10} \\ \\ \sf \mapsto E = 4.2 \times {(10)}^{10} \: \: N/c$

$\therefore \sf \underline{The \: dipole \: field \: at \: an \: axial \: point \: and \: equitorial \: point \: are \: 4.2 × {(10)}^{10} \: N/c \: and \: 2.1 × {(10)}^{10} \: N/c}$