Question

An electric dipole is made of two charges of equal magnitudes and opposite signs. The positive charge, q=1.0 μC, is located at the point (x, y, z) = (0.00 cm, 1.0 cm, 0.00 cm), while the negative charge is located at the point (x, y, z) = (0.00 cm, −1.0 cm, 0.00 cm). How much work will be done by an electric field E⃗ =3.0×106 N/C) i^ to bring the dipole to its stable equilibrium position?

Answer 1

Answer:

work done by electric field  is 0.06 J

Explanation:

Given data:

Two point charge is $+ 1\mu C and -1 \mu C$

0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)

-1 charge positioned is (0 cm , -1 cm, 0.00 cm)

E = 3.0\times 10^6 N/C

From above information, the distance between  given two charges d = 2 cm

then d = 0.02m

 work needed is W = q E d

$W = 1.0 \times 10^{-6} \times 3.0 \times 10^6 \times 0.02$

W = 0.06 J  

Therefore work done by electric field  is 0.06 J