An electric dipole is constructed using equal and opposite charges of magnitude 10 `muC` and placing them at a separation 1 mm. There is a point P at a distance 1 m from the centre of dipole. (i) Calculate electric field intensity and potential at point P. (ii) What will be the answers to the part. (i) if electric dipole is turned through right angle from its existing orientation?

Explanation

(i) Potential at any point on the axial line of a short dipole of dipole moment p at a distance r from its centre is given by

Electric field at this position will be given by
`E=(1)/(4pi epsilon_(0))*(p)/(r^(3))`
Here, `p=q xx 2l=10xx10^(-6)xx1xx10^(-3)Cm`
`=10xx10^(-9)Cm`
`therefore E=(9xx10^(9)xx10xx10^(-9))/((1)^(3))=90NC^(-1)`
Electric potential at this position (any point on equatorial plane) will be equal to zero.
(ii) Electric field at this position will be given by:

`E=(1)/(4pi epsilon_(0))*(2p)/(r^(3))`
`=(9xx10^(9)xx2xx10xx10^(-9))/((1)^(3))`
`=180NC^(-1)`
Electric potential at this point will be given by:
`V=(1)/(4pi epsilon_(0))*(p)/(r^(2))`
`=(9xx10^(9)xx10xx10^(-9))/((1)^(2))`
`=90V`