An electric dipole is placed as shown in the figure. [figure] The electric potential (in 102 V) at point P due to the dipole is ( 0 = permittivity of free space and 14 0 = K) :
- ( 58 ) qK
- ( 85 ) qK
- ( 83 ) qK
- ( 38 ) qK
Answer
(4) ( 38 ) qK
Solution
P lies on the axis, 5 cm from the centre O.
Distance from +q: r1 = 5 - 3 = 2 cm = 0.02 m
Distance from -q: r2 = 5 + 3 = 8 cm = 0.08 m
V = Kq ( 1r1 - 1r2 )
V = Kq ( 10.02 - 10.08 )
V = Kq(50 - 12.5) = 37.5 qK
V = ( 38 ) qK 102 V
Answer: (4)