Please give an explanation An electric dipole consisting of charges of magnitude

Question

Please give an explanation

An electric dipole consisting of charges of magnitude 4.70 nC separated by 6.20 m is in an electric field of strength 1350 N/C (a) What is the magnitude of the electric dipole moment? 2.914e-14 C-m (b) What is the difference between potential energies for dipole orientations parallel to and antiparallel to E? 3.52e-9 Do you realize that the dipole moment magnitude equals the product of separation distance and the charge magnitude at either end? (Don't use the charge twice.) Do you realize that "antiparallel" means opposite? Did you plug into the potential energy equation?

Explanation / Answer

Given q = 4.7 nC

distance d = 6.20 µm

electric field E = 1350 N/C

a) The electric dipole is

p = q * d

p = 4.7 * 10^-9 * 6.20 * 10^-6

p = 2.914 * 10^-14 C-m

b)

the pontential energy of a dipole in an electric field is given by

U = -p * E * cos(theta)

For parallel orientation theta = 0 so

U1 = - p * E * cos(0)

U1 = - 2.914 * 10^-14 * 1350 * 1

U1 = - 3.93 * 10^-11 J

For Anti-parallel orientation theta = 180

U2 = - p * E * cos(180)

U2 = - 2.914 * 10^-14 * 1350 * (-1)

U2 = + 3.93 * 10^-11 J

The difference in energies is

U = U2 - U1

U = + 3.93 * 10^-11 - (- 3.93 * 10^-11)

U = 7.86 * 10^-11 J------------------------------------------Answer

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---