Contact

For Study plan details

10:00 AM to 7:00 PM IST all days.

Franchisee/Partner Enquiry (South)

Franchisee/Partner Enquiry (North, West & East)

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

9372462318 / 9987178554

Mon to Sat - 10 AM to 7 PM

# Find the potential of the field created by the dipole at point A, located at a distance  r=0.5 m, in a direction at an angle 300,   relative to the electric moment of the dipole. Medium is water. The dipole is formed by charges q=2·10-7 C, that   located at a distance l=0.5 cm.

Asked by Zithinmuppalla 10th March 2021, 3:09 PM
Potential of dipole is given as

where p is electric dipole moment, θ is angle made by line joining dipole to the point where potential is required
with direction of dipole moment, εr is dielectric constant , εo is the permeability of free space and r is the distance
between dipole and the point where potential is required.

electric dipole moment  p = ( q d ) = 2 × 10-7 × 0.5 × 10-2 = 10-9 C m

dielectric constant of water εr = 80.4

Coulomb constant = 1/ ( 4πεo ) = 9 × 109 N m2 C-2

V =  ( 9 × 109 × 10-9 × cos300 ) / ( 80.4 × 0.5 × 0.5 ) = 0.224 Volt
Answered by Expert 10th March 2021, 5:07 PM
• 1
• 2
• 3
• 4
• 5
• 6
• 7
• 8
• 9
• 10

You have rated this answer /10

### Free related questions

15th February 2019, 10:16 AM
27th March 2019, 3:44 PM
RELATED STUDY RESOURCES :