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`color{blue}{star}` EQUIPOTENTIAL SURFACES


`color{blue} ✍️` An equipotential surface is a surface with a constant value of potential at all points on the surface.

`color{blue} ✍️` For a single charge `q,` the potential is given by Eq. (2.8):

`color{blue} {V = 1/(4piepsilon_0) q/r}`

`color{blue} ✍️` This shows that `V` is a constant if `r` is constant . Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge.

`color{blue} ✍️` Now the electric field lines for a single charge `q` are radial lines starting from or ending at the charge, depending on whether `q` is positive or negative.

`color{blue} ✍️` Clearly, the electric field at every point is normal to the equipotential surface passing through that point. This is true in general for any charge configuration, equipotential surface through a point is normal to the electric field at that point.

`color{blue} ✍️` The proof of this statement is simple. If the field were not normal to the equipotential surface, it would have non-zero component along the surface.

`color{blue} ✍️` To move a unit test charge against the direction of the component of the field, work would have to be done.

`color{blue} ✍️` But this is in contradiction to the definition of an equipotential surface: there is no potential difference between any two points on the surface and no work is required to move a test charge on the surface.

`color{blue} ✍️` The electric field must, therefore, be normal to the equipotential surface at every point. Equipotential surfaces offer an alternative visual picture in addition to the picture of electric field lines around a charge configuration.

`color{blue} ✍️` For a uniform electric field E, say, along the x -axis, the equipotential surfaces are planes normal to the x -axis, i.e., planes parallel to the y-z plane (Fig. 2.10). Equipotential surfaces for (a) a dipole and (b) two identical positive charges are shown in Fig. 2.11.

`color{blue}ulbb "Relation between field and potential"`

`color{blue} ✍️` Consider two closely spaced equipotential surfaces A and B (Fig. 2.12) with potential values `V` and `V + δV,` where `δV` is the change in `V` in the direction of the electric field `E.`

`color{blue} ✍️` Let `P` be a point on the surface `B`. `δl` is the perpendicular distance of the surface A from `P`. Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. The work done in this process is `|E|δ l.`

`color{blue} ✍️` This work equals the potential difference `V_A–V_B`


`|E|δ l = V−(V +δV)= –δV`

i.e., `color{blue} {|E|= (δV)/(δ l)}` .................... 2.20

Since `δV` is negative, `δV = – |δV|.` we can rewrite Eq (2.20) as

`color{blue}{|E|= (δV)/(δ l) = + (|δV|)/(δ l)}`


`color{blue} ✍️` We thus arrive at two important conclusions concerning the relation between electric field and potential:

`color{blue} ✍️` Electric field is in the direction in which the potential decreases steepest.

`color{blue} ✍️` Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.


`color{blue} ✍️` Consider first the simple case of two charges `q_1` and `q_2` with position vector `r_1` and `r_2` relative to some origin.

`color{blue} ✍️` Let us calculate the work done (externally) in building up this configuration. This means that we consider the charges `q_1` and `q_2` initially at infinity and determine the work done by an external agency to bring the charges to the given locations.

`color{blue} ✍️` Suppose, first the charge `q_1` is brought from infinity to the point `r_1`.

`color{blue} ✍️` There is no external field against which work needs to be done, so work done in bringing `q_1` from infinity to `r_1` is zero. This charge produces a potential in space given by

`color{green} {V = 1/(4piepsilon_0) (q_1)/(r_(1P)}`

`color{blue} ✍️` where `r_(1P)` is the distance of a point P in space from the location of `q_1.` From the definition of potential, work done in bringing charge `q_2` from infinity to the point `r_2` is `q_2` times the potential at `r_2` due to `q_1:`

work done on `color{green} {q_2= 1/(4piepsilon_0) (q_1q_2)/(r_12)}`

`color{blue} ✍️` where `r_(12)` is the distance between points 1 and 2. Since electrostatic force is conservative, this work gets stored in the form of potential energy of the system. Thus, the potential energy of a system of two charges `q_1` and `q_2` is

`U = 1/(4piepsilon_0) (q_1q_2)/(r_12)` ........................ 2.22

`color{blue} ✍️` Obviously, if `q_2` was brought first to its present location and `q_1` brought later, the potential energy `U` would be the same.

`color{blue} ✍️` More generally, the potential energy expression, Eq. (2.22), is unaltered whatever way the charges are brought to the specified locations, because of path-independence of work for electrostatic force.

`color{blue} ✍️` Equation (2.22) is true for any sign of `q_1` and `q_2`. If `q_1q_2 > 0`, potential energy is positive.

`color{blue} ✍️` This is as expected, since for like charges `(q_1q_2 > 0),` electrostatic force is repulsive and a positive amount of work is needed to be done against this force to bring the charges from infinity to a finite distance apart. For unlike charges `(q_1 q_2 < 0),` the electrostatic force is attractive.

`color{blue} ✍️` In that case, a positive amount of work is needed against this force to take the charges from the given location to infinity.

`color{blue} ✍️` In other words, a negative amount of work is needed for the reverse path (from infinity to the present locations), so the potential energy is negative. Equation (2.22) is easily generalised for a system of any number of point charges.

`color{blue} ✍️` Let us calculate the potential energy of a system of three charges `q_1, q_2 and q_3` located at `r_1, r_2, r_3`, respectively.

`color{blue} ✍️` To bring `q_1` first from infinity to `r_1`, no work is required. Next we bring `q_2` from infinity to `r_2`. As before, work done in this step is

`q_2 V_1 (r_2) = 1/(4piepsilon_0)(q_1q_2)/(r_12)` ..................2.23

`color{blue} ✍️` The charges `q_1` and `q_2` produce a potential, which at any point P is given by

`V(1,2) = 1/(4piepsilon_0) ((q_1)/(r_(1P)) +(q_1)/(r_(2P)))` .......................... 2.24

`color{blue} ✍️` Work done next in bringing `q_3` from infinity to the point `r_3` is `q_3` times `V_(1, 2)` at `r_3`

`q_2 V_(1,2) (r_2) = 1/(4piepsilon_0) ((q_1q_3)/(r_13)+(q_2q_3)/(r_23))` ............................. 2.25

`color{blue} ✍️` The total work done in assembling the charges at the given locations is obtained by adding the work done in different steps [Eq. (2.23) and Eq. (2.25)],

`U = 1/ (4 pi epsi_0 ) ( ( q_1q_2)/r_(12) + ( q_1q_3)/r_(13) + (q_2q_3)/r_23 )`


`color{blue} ✍️` Again, because of the conservative nature of the electrostatic force (or equivalently, the path independence of work done), the final expression for U, Eq. (2.26), is independent of the manner in which the configuration is assembled.

`color{blue} ✍️` The potential energy is characteristic of the present state of configuration, and not the way the state is achieved.
Q 3174680556

Four charges are arranged at the corners of a square `ABCD` of side d, as shown in Fig. 2.15.(a) Find the work required to put together this arrangement. (b) A charge `q_0` is brought to the centre E of the square, the four charges being held fixed at its corners. How much extra work is needed to do this?
Class 12 Chapter 2 Example 4

(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose,first the charge +q is brought to A, and then the charges –q, +q, and –q are brought to B, C and D, respectively. The total work needed can be calculated in steps:
(i) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero.
(ii) Work needed to bring –q to B when +q is at A. This is given by
(charge at B) × (electrostatic potential at B due to charge +q at A

`color{blue}{= - q xx(q)/(4piepsilon_0d) = - (q^2)/(4piepsilon_0d)}`
iii) Work needed to bring charge +q to C when +q is at A and –q is at
B. This is given by (charge at C) × (potential at C due to charges at A and B)

`= +q[(+q)/(4piepsilon_0dsqrt2)+(-q)/(4piepsilon_0d)]`

`= (-q^2)/(4piepsilon_0d)(1 - 1/sqrt2)`

(iv) Work needed to bring –q to D when +q at A,–q at B, and +q at C. This is given by (charge at D) × (potential at D due to charges at A, B and C)

`= - q ((+q)/(4piepsilon_0d)+(-q)/(4piepsilon_0dsqrt2)+(q)/(4piepsilon_0d))`

`= (-q^2)/(4piepsilon_0d) (2-1/sqrt2)`

Add the work done in steps (i), (ii), (iii) and (iv). The total work
required is

`= (-q)/(4piepsilon_0d) {(0) +(1) +(1-1/sqrt2) + (2 - 1/sqrt2)}`

`= (-q^2)/(4piepsilon_0d)(4-sqrt2)`

The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges. (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.) (b) The extra work necessary to bring a charge `q_0` to the point E when the four charges are at A, B, C and D is `q_0` × (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D. Hence no work is required to bring any charge to point E.