Q83P - Fundamentals Of Physics

Question

An electric dipole with dipole moment $\overrightarrow{\mathbf{p}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.00}\widehat{\mathbf{i}}\mathbf{+}\mathbf{4}\mathbf{.00}\widehat{\mathbf{j}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{.24}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{30}\mathbf{ }}\mathbf{ }\mathbf{\text{C.m}}\mathbf{)}$is in an electric field$\overrightarrow{\mathbf{E}}\mathbf{=}\mathbf{\left(}\mathbf{4000}\mathbf{\text{ N/C}}\mathbf{\right)}\widehat{\mathbf{i}}$ (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 is$\overrightarrow{\mathbf{p}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{.00}\widehat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.00}\widehat{\mathbf{j}}\mathbf{)}\mathbf{(}\mathbf{1}\mathbf{.24}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{30}\mathbf{ }}\mathbf{\text{ C.m}}\mathbf{)}$ how much work is done by the agent?

Step-by-Step Solution

Answer

a) The potential energy of the electric dipole is$- 1.49\times {10}^{-26 }\text{J}$ .

b) The torque acting on it is$(- 1.98\times {10}^{-26 }\text{N}\cdot \text{m})\widehat{k}$ .

c) The work done by the agent is $3.47\times {10}^{-26 }\text{J}$.  

 

Step 1: The given data

a) Initial electric dipole moment, $\overrightarrow{p}=(3.00\widehat{i}+4.00\widehat{j})(1.24\times {10}^{-30 } \text{C}\cdot \text{m})$

b) Final electric dipole moment,$\overrightarrow{p}=(-4.00\widehat{i}+3.00\widehat{j})(1.24\times {10}^{-30 } \text{C}\cdot \text{m})$

c) Strength of electric field, $\overrightarrow{E}=\left(4000\text{ N/C}\right)\widehat{i}$

Step 2: 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 is$\mathit{p}\mathbf{=}\mathit{q}\mathit{d}$ , 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)$U\left(\theta \right)=-\overrightarrow{p}.\overrightarrow{E}=-pEcos\theta$

Therefore, if the initial angle between p and E is and the final angle is$\theta$ , then the change in potential energy is given as:$\Delta U= U\left(\theta \right)- U_o\left(\theta \right)$ (ii)

Net torque acting on a dipole placed in an Electric field is given by:$\overrightarrow{\tau }=\overrightarrow{p}\times \overrightarrow{E}$   (iii)

 

Step 3: 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:

$\begin{array}{c}U=-\left[(3.00\widehat{i}+4.00\widehat{j})(1.24\times {10}^{-30 }\text{ C×m})\right].[\left(4000\text{ N/C}\right)\widehat{i} ]\\ = - 1.49\times {10}^{-26}\text{J}\end{array}$

Hence, the value of the potential energy is.$- 1.49\times {10}^{-26 }J$

 

Step 4: b) Calculation of the torque acting on the dipole

Using the given data in equation (ii) (and the facts that  $\widehat{i} \times \widehat{i}$= o and  $\widehat{i}\times \widehat{j}$= $\widehat{k}$  ), the torque acting on the dipole is given as follows:

$\begin{array}{c}\overrightarrow{\tau }=\left[(3.00\widehat{i}+4.00\widehat{j})(1.24\times {10}^{-30 } \text{C}\cdot \text{m})\right]\times [\left(4000\text{ N/C}\right)\widehat{i} ]\\ =(- 1.98\times {10}^{-26 }\text{N}\cdot \text{m})\widehat{k}\end{array}$

Hence, the value of the torque is$(- 1.98\times {10}^{-26 }\text{N}\cdot \text{m})\widehat{k}$.

 

Step 5: 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:

$\begin{array}{c}W=\Delta U\\ =|{\overrightarrow{p}}_i-{\overrightarrow{p}}_f |\cdot \overrightarrow{E}\\ = \left[\{(3.00\widehat{i}+4.00\widehat{j})-(-4.00\widehat{i}+3.00\widehat{j})\}(1.24\times {10}^{-30 } \text{C}\cdot \text{m})\right]\cdot [\left(4000\text{ N/C}\right)\widehat{i} ]\\ = 3.47\times {10}^{-26 }\text{J}\end{array}$

Hence, the value of the work done is.$3.47\times {10}^{-26 }\text{J}$