`color{blue}{star}` POTENTIAL ENERGY IN AN EXTERNAL FIELD

`color{blue}{star}` POTENTIAL ENERGY OF A DIPOLE AN EXTERNAL FILED

`color{blue}{star}` POTENTIAL ENERGY OF A DIPOLE AN EXTERNAL FILED

`color{brown} bbul"Potential energy of a single charge"`

`color{blue} ✍️` In Section 2.7, the source of the electric field was specified – the charges and their locations - and the potential energy of the system of those charges was determined.

`color{blue} ✍️` The main difference is that we are now concerned with the potential energy of a charge (or charges) in an external field.

`color{blue} ✍️` The external field `E` is not produced by the given charge(s) whose potential energy we wish to calculate. E is produced by sources external to the given charge(s).

`color{blue} ✍️` The external sources may be known, but often they are unknown or unspecified; what is specified is the electric field E or the electrostatic potential V due to the external sources.

`color{blue} ✍️` We assume that the charge `q` does not significantly affect the sources producing the external field. This is true if `q` is very small, or the external sources are held fixed by other unspecified forces.

`color{blue} ✍️` Even if q is finite, its influence on the external sources may still be ignored in the situation when very strong sources far away at infinity produce a finite field E in the region of interest.

`color{blue} ✍️` Note again that we are interested in determining the potential energy of a given charge `q` (and later, a system of charges) in the external field; we are not interested in the potential energy of the sources producing the external electric field.

`color{blue} ✍️` The external electric field `E` and the corresponding external potential `V` may vary from point to point. By definition, `V` at a point `P` is the work done in bringing a unit positive charge from infinity to the point `P.`

(We continue to take potential at infinity to be zero.) Thus, work done in bringing a charge `q` from infinity to the point `P` in the external field is qV. This work is stored in the form of potential energy of `q`. If the point P has position vector `r` relative to some origin, we can write: Potential energy of `q` at `r` in an external field

`= qV(r)`.................... 2.27

`color{blue} ✍️` where `V(r)` is the external potential at the point `r`.

`color{blue} ✍️` Thus, if an electron with charge `color{fuchsia}(q = e = 1.6c×10^(–19) C)` is accelerated by a potential difference of `ΔV = 1` volt, it would gain energy of `qΔV = 1.6 × 10^(–19)J`.

`color{blue} ✍️` This unit of energy is defined as 1 electron volt or `1eV,` i.e., `1 eV=1.6 × 10^(–19)J.`

`color{blue} ✍️` The units based on `eV` are most commonly used in atomic, nuclear and particle physics,

`(1 keV = 10^3eV = 1.6 × 10^(–16)J, 1 MeV = 10^(6)eV = 1.6 × 10^(–13)J, 1 GeV = 10^9eV = 1.6 × 10^(–10)J` and `1 TeV = 10^(12)eV = 1.6 × 10^(–7)J.`

`color{blue}(bbul"Potential energy of a system of two charges in an external field")`

`color{blue} ✍️` Next, we ask what is the potential energy of a system of two charges `q_1` and `q_2` located at `r_1`and `r_2`, respectively, in an external field? First, we calculate the work done in bringing the charge `q_1` from infinity to `r_1`.

`color{blue} ✍️` Work done in this step is `q_1 V(r_1)`, using Eq. (2.27). Next, we consider the work done in bringing `q_2` to `r_2`.

`color{blue} ✍️` In this step, work is done not only against the external field E but also against the field due to `q_1`.

`color{blue} ✍️` Work done on `q_2` against the external field

`color{green} {= q_2V((r_2))}`

`color{blue} ✍️` Work done on `q_2` against the field due to `q_1`

`color{green}( = (q_1q_2)/(4piepsilon_0r_12))`

where `r_(12)` is the distance between `q_1` and `q_2`. We have made use of Eqs. (2.27) and (2.22). By the superposition principle for fields, we add up the work done on `q_2` against the two fields (E and that due to `q_1`): Work done in bringing `q_2` to `r_2`

`color{green}{= (q_2V(r_2) +(q_1)(q_2)/(4piepsilon_0r_12))}` ...................2.28

Thus, Potential energy of the system = the total work done in assembling the configuration

`color{green}(= q_1 V(r_1) +q_2V(r_2) + (q_1q_2)/(4piepsilon_0r_12))` ..............2.28

`color{blue} ✍️` In Section 2.7, the source of the electric field was specified – the charges and their locations - and the potential energy of the system of those charges was determined.

`color{blue} ✍️` The main difference is that we are now concerned with the potential energy of a charge (or charges) in an external field.

`color{blue} ✍️` The external field `E` is not produced by the given charge(s) whose potential energy we wish to calculate. E is produced by sources external to the given charge(s).

`color{blue} ✍️` The external sources may be known, but often they are unknown or unspecified; what is specified is the electric field E or the electrostatic potential V due to the external sources.

`color{blue} ✍️` We assume that the charge `q` does not significantly affect the sources producing the external field. This is true if `q` is very small, or the external sources are held fixed by other unspecified forces.

`color{blue} ✍️` Even if q is finite, its influence on the external sources may still be ignored in the situation when very strong sources far away at infinity produce a finite field E in the region of interest.

`color{blue} ✍️` Note again that we are interested in determining the potential energy of a given charge `q` (and later, a system of charges) in the external field; we are not interested in the potential energy of the sources producing the external electric field.

`color{blue} ✍️` The external electric field `E` and the corresponding external potential `V` may vary from point to point. By definition, `V` at a point `P` is the work done in bringing a unit positive charge from infinity to the point `P.`

(We continue to take potential at infinity to be zero.) Thus, work done in bringing a charge `q` from infinity to the point `P` in the external field is qV. This work is stored in the form of potential energy of `q`. If the point P has position vector `r` relative to some origin, we can write: Potential energy of `q` at `r` in an external field

`= qV(r)`.................... 2.27

`color{blue} ✍️` where `V(r)` is the external potential at the point `r`.

`color{blue} ✍️` Thus, if an electron with charge `color{fuchsia}(q = e = 1.6c×10^(–19) C)` is accelerated by a potential difference of `ΔV = 1` volt, it would gain energy of `qΔV = 1.6 × 10^(–19)J`.

`color{blue} ✍️` This unit of energy is defined as 1 electron volt or `1eV,` i.e., `1 eV=1.6 × 10^(–19)J.`

`color{blue} ✍️` The units based on `eV` are most commonly used in atomic, nuclear and particle physics,

`(1 keV = 10^3eV = 1.6 × 10^(–16)J, 1 MeV = 10^(6)eV = 1.6 × 10^(–13)J, 1 GeV = 10^9eV = 1.6 × 10^(–10)J` and `1 TeV = 10^(12)eV = 1.6 × 10^(–7)J.`

`color{blue}(bbul"Potential energy of a system of two charges in an external field")`

`color{blue} ✍️` Next, we ask what is the potential energy of a system of two charges `q_1` and `q_2` located at `r_1`and `r_2`, respectively, in an external field? First, we calculate the work done in bringing the charge `q_1` from infinity to `r_1`.

`color{blue} ✍️` Work done in this step is `q_1 V(r_1)`, using Eq. (2.27). Next, we consider the work done in bringing `q_2` to `r_2`.

`color{blue} ✍️` In this step, work is done not only against the external field E but also against the field due to `q_1`.

`color{blue} ✍️` Work done on `q_2` against the external field

`color{green} {= q_2V((r_2))}`

`color{blue} ✍️` Work done on `q_2` against the field due to `q_1`

`color{green}( = (q_1q_2)/(4piepsilon_0r_12))`

where `r_(12)` is the distance between `q_1` and `q_2`. We have made use of Eqs. (2.27) and (2.22). By the superposition principle for fields, we add up the work done on `q_2` against the two fields (E and that due to `q_1`): Work done in bringing `q_2` to `r_2`

`color{green}{= (q_2V(r_2) +(q_1)(q_2)/(4piepsilon_0r_12))}` ...................2.28

Thus, Potential energy of the system = the total work done in assembling the configuration

`color{green}(= q_1 V(r_1) +q_2V(r_2) + (q_1q_2)/(4piepsilon_0r_12))` ..............2.28

Q 3134780652

(a) Determine the electrostatic potential energy of a system consisting of two charges `7 μC` and `–2 μC` (and with no external field) placed at `(–9 cm, 0, 0)` and `(9 cm, 0, 0)` respectively.

(b) How much work is required to separate the two charges infinitely away from each other?

(c) Suppose that the same system of charges is now placed in an external electric field `E = A (1/r 2); A = 9 × 105 C m–2`. What would the electrostatic energy of the configuration be?

Class 12 Chapter example 5

(b) How much work is required to separate the two charges infinitely away from each other?

(c) Suppose that the same system of charges is now placed in an external electric field `E = A (1/r 2); A = 9 × 105 C m–2`. What would the electrostatic energy of the configuration be?

Class 12 Chapter example 5

`(a) color{blue}{U = 1/(4piepsilon_0) (q_1q_2)/(r) }= 9xx10^9 xx7xx(-2)xx10^(-12)/(0.18) = - 0.7 J`

`(b) W = U_2 – U_1 = 0 – U = 0 – (–0.7) = 0.7 J.`

(c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,

`(b) W = U_2 - U_1 = 0 - (-0.7) = 0.7 J`

c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,

`q_1 V (r_1) +q_2V(r_2) = A(7muC)/(0.09m) +A(-2muC)/(0.009m)`

and the net electrostatic energy is

`q_1V(r_1)+q_2V(r_2) +(q_1q_2)/(4piepsilon_0r_12) = A(7muC)/(0.09m) +A(-2muC)/(0.009m)- 0.7J`

`= 70-20-0.7 = 49.3J`

`color{blue} ✍️` Consider a dipole with charges `q_1 = +q` and `q_2 = –q` placed in a uniform electric field E, as shown in Fig. 2.16.

`color{blue} ✍️` As seen in the last chapter, in a uniform electric field, the dipole experiences no net force; but experiences a torque τ given by

`τ = p×E` ....................2.30

`color{blue} ✍️` which will tend to rotate it (unless p is parallel or antiparallel to `E`).

`color{blue} ✍️` Suppose an external torque `τ_"ext"` is applied in such a manner that it just neutralizes this torque and rotates it in the plane of paper from angle θ0 to angle θ1 at an infinitesimal angular speed and without angular acceleration.

`color{blue} ✍️` The amount of work done by the external torque will be given by

`color{blue} ✍️` This work is stored as the potential energy of the system. We can then associate potential energy `U(θ )` with an inclination `θ` of the dipole.

`color{blue} ✍️` Similar to other potential energies, there is a freedom in choosing the angle where the potential energy U is taken to be zero.

`color{blue} ✍️` A natural choice is to take `color{fuchsia}(θ_0 = π / 2).` (Αn explanation for it is provided towards the end of discussion.) We can then write,

`color{green}(U (theta)= pE (cos \ pi/2- cos theta) = - pE cos theta = - P*E)` ..................2.32

`color{blue} ✍️` This expression can alternately be understood also from Eq. (2.29). We apply Eq. (2.29) to the present system of two charges `+q` and `–q.` The potential energy expression then reads

`color{green}{U' (theta)=q [V(r_1)-V(r_2)] - (q^2)/(4piepsilon_0xx2a)}`

Here, `r_1` and `r_2` denote the position vectors of `+q` and `–q`. Now, the potential difference between positions `r_1` and `r_2` equals the work done in bringing a unit positive charge against field from `r_2` to `r_1.` The displacement parallel to the force is `2a cosθ`. Thus,

`color{fuchsia}([V(r_1)–V (r_2)] = –E × 2a cosθ)` . We thus obtain,

`color{green}{U' (theta)=q [V(r_1)-V(r_2)] - (q^2)/(4piepsilon_0xx2a) = - P*E - (q^2)/(4piepsilon_0)xx2a)` ..................2.34

`color{blue} ✍️` We note that `U′ (θ )` differs from `U(θ )` by a quantity which is just a constant for a given dipole.

`color{blue} ✍️` Since a constant is insignificant for potential energy, we can drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32).

`color{blue} ✍️` We can now understand why we took `θ=π//2.` In this case, the work done against the external field `E` in bringing `+q` and `– q` are equal and opposite and cancel out, i.e., `color{fuchsia}{q [V (r_1) – V (r_2)]=0.}`

`color{blue} ✍️` As seen in the last chapter, in a uniform electric field, the dipole experiences no net force; but experiences a torque τ given by

`τ = p×E` ....................2.30

`color{blue} ✍️` which will tend to rotate it (unless p is parallel or antiparallel to `E`).

`color{blue} ✍️` Suppose an external torque `τ_"ext"` is applied in such a manner that it just neutralizes this torque and rotates it in the plane of paper from angle θ0 to angle θ1 at an infinitesimal angular speed and without angular acceleration.

`color{blue} ✍️` The amount of work done by the external torque will be given by

`color{green}(W = int_(theta_0)^(theta_1) τ_(ext)(theta)d theta = int_(theta_0)^(theta_1) PE sin theta d theta)`

`color{green}{= pE (costheta_0 - costheta_1)}`

`color{blue} ✍️` This work is stored as the potential energy of the system. We can then associate potential energy `U(θ )` with an inclination `θ` of the dipole.

`color{blue} ✍️` Similar to other potential energies, there is a freedom in choosing the angle where the potential energy U is taken to be zero.

`color{blue} ✍️` A natural choice is to take `color{fuchsia}(θ_0 = π / 2).` (Αn explanation for it is provided towards the end of discussion.) We can then write,

`color{green}(U (theta)= pE (cos \ pi/2- cos theta) = - pE cos theta = - P*E)` ..................2.32

`color{blue} ✍️` This expression can alternately be understood also from Eq. (2.29). We apply Eq. (2.29) to the present system of two charges `+q` and `–q.` The potential energy expression then reads

`color{green}{U' (theta)=q [V(r_1)-V(r_2)] - (q^2)/(4piepsilon_0xx2a)}`

Here, `r_1` and `r_2` denote the position vectors of `+q` and `–q`. Now, the potential difference between positions `r_1` and `r_2` equals the work done in bringing a unit positive charge against field from `r_2` to `r_1.` The displacement parallel to the force is `2a cosθ`. Thus,

`color{fuchsia}([V(r_1)–V (r_2)] = –E × 2a cosθ)` . We thus obtain,

`color{green}{U' (theta)=q [V(r_1)-V(r_2)] - (q^2)/(4piepsilon_0xx2a) = - P*E - (q^2)/(4piepsilon_0)xx2a)` ..................2.34

`color{blue} ✍️` We note that `U′ (θ )` differs from `U(θ )` by a quantity which is just a constant for a given dipole.

`color{blue} ✍️` Since a constant is insignificant for potential energy, we can drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32).

`color{blue} ✍️` We can now understand why we took `θ=π//2.` In this case, the work done against the external field `E` in bringing `+q` and `– q` are equal and opposite and cancel out, i.e., `color{fuchsia}{q [V (r_1) – V (r_2)]=0.}`

Q 3174780656

A molecule of a substance has a permanent electric dipole moment of magnitude `10^(–29)` C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude `106 V m^(–1.)` The direction of the field is suddenly changed by an angle of `60º`. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume `100%` polarisation of the sample.

Class 12 Chapter example 6

Class 12 Chapter example 6

Here, dipole moment of each molecules `= 10^(–29)` C m

As 1 mole of the substance contains `6 × 10^23` molecules,

total dipole moment of all the molecules, `p = 6 × 10^23 × 10^(–29) C m = 6 × 10^(–6)

C m`

Initial potential energy, `Ui = –pE cos θ = –6×10^(–6)×10^6 cos 0° = –6 J`

Final potential energy (when `θ = 60°`), Uf `= –6 × 10^(–6) × 10^6 cos 60° = –3 J`

Change in potential energy `= –3 J – (–6J) = 3 J`

So, there is loss in potential energy. This must be the energy released

by the substance in the form of heat in aligning its dipoles.