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An electric dipole with dipole momentp→=(3.00i^+4.00j^)(1.24×10−30C.m)is in an electric fieldE→=(4000N/C)i^(a) What is the potentialenergy of the electric dipole? (b) What is the torque acting on it?(c) If an external agent turns the dipole until its electric dipole moment isp→=(−4.00i^+3.00j^)(1.24×10−30C.m)how much work is done by the agent?

Short Answer

Expert verified

a) The potential energy of the electric dipole is−1.49×10−26J .

b) The torque acting on it is(−1.98×10−26N⋅m)k^ .

c) The work done by the agent is 3.47×10−26J.

Step by step solution

01

The given data

a) Initial electric dipole moment,p→=(3.00i^+4.00j^)(1.24×10−30C⋅m)

b) Final electric dipole moment,p→=(−4.00i^+3.00j^)(1.24×10−30C⋅m)

c) Strength of electric field,E→=(4000N/C)i^

02

Understanding the concept of the electric field and the energy

The potential energy of the electric dipole placed in an electric field depends on its orientation relative to the electric field.The magnitude of the electric dipole moment isp=qd, where q is the magnitude of the charge, and d is the separation between the two charges. Using the concept of the potential energy of the dipole, we can get the required values. Again, using this value, we can get the value of torque acting on the respective dipole.

Formulae:

When placed in an electric field, the potential energy of the dipole,

(i)U(θ)=−p→.E→=−pEcosθ

Therefore, if the initial angle between p and E isand the final angle isθ , then the change in potential energy is given as:ΔU=U(θ)−Uo(θ) (ii)

Net torque acting on a dipole placed in an Electric field is given by:τ→=p→×E→ (iii)

03

a) Calculation of the potential energy of the dipole

The potential energy of the electric dipole placed in an electric field depends on its orientation relative to the electric field. The field causes a torque that tends to align the dipole with the field.

When placed in an electric field E, the potential energy of the dipole p is given by equation (i) as follows:

U=−[(3.00i^+4.00j^)(1.24×10−30°ä׳¾)].[(4000N/C)i^]=−1.49×10-26J

Hence, the value of the potential energy is.−1.49×10−26J

04

b) Calculation of the torque acting on the dipole

Using the given data in equation (ii) (and the facts thati^×i^= o andi^×j^=k^ ), the torque acting on the dipole is given as follows:

τ→=[(3.00i^+4.00j^)(1.24×10−30C⋅m)]×[(4000N/C)i^]=(−1.98×10−26N⋅m)k^

Hence, the value of the torque is(−1.98×10−26N⋅m)k^.

05

c) Calculation of the work done

The work done by the agent is equal to the change in the potential energy of the dipole. Thus, using equation (i) in equation (ii), the work done is given as:

W=ΔU=|p→i−p→f|⋅E→=[{(3.00i^+4.00j^)−(−4.00i^+3.00j^)}(1.24×10−30C⋅m)]⋅[(4000N/C)i^]=3.47×10−26J

Hence, the value of the work done is.3.47×10−26J

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