Question

• An electric dipole has magnitude
q = 2.4 micro coulomb and separation
2.5 x 10 –3 cm. The electric field at
distance 0.30 m on axial point is
Answer
A. 4 x 1010 N/C
B.O 4 x 1011 N/C
C.O 4/3 x 1010 N/C
D.O 4/3 x 1011 N/C​

Answer 1

Given:

An electric dipole has magnitude q = 2.4 micro coulomb and separation 2.5 x 10 –3 cm.

To find:

Field intensity on axis at a distance of 0.3 metres.

Calculation:

Since, the separation distance between the charges (in the dipole) is much less as compared to 0.3 metres, we can consider the dipole to be a short dipole

For a short dipole:

• Field intensity on an axial position is given as follows:

$\boxed{ \bf \: E = \dfrac{2kp}{ {x}^{3} } }$

• 'k' is Coulomb's Constant
• 'p' is Dipole moment

Putting all the values in SI UNITS:

$\rm \implies\: E = \dfrac{2 \times (9 \times {10}^{9}) \times (q \times d)}{ {(0.3)}^{3} }$

$\rm \implies\: E = \dfrac{2 \times (9 \times {10}^{9}) \times (2.4 \times {10}^{ - 6} \times 2.5 \times {10}^{ - 5} )}{ {(0.3)}^{3} }$

$\rm \implies\: E = \dfrac{108 \times {10}^{ - 2}}{0.027}$

$\rm \implies\: E = 4000 \times {10}^{ - 2}$

$\rm \implies\: E = 40 \: N/C$

So, field intensity at that position is 40 N/C.