Explanation :

If a test charge q₀ is moved on an equipotential surface of potential V, the electric potential energy U = q₀V remains constant. Because U does not change as q₀ is moved, the work done by the electric field on q₀ must be zero.

If  is the electric field on the surface,

dW = $\vec{F}.\vec{dx}$  = $q_0\vec{F}.\vec{dx}$   = $q_0\vec{E}cos θ$ = 0

∴ cos θ = 0  or  θ = 90⁰

Hence, the electric force $q_0\vec{E}$  is always perpendicular to the displacement of a charge moving on an equilateral surface. Thus, electric field lines and equipotential surfaces are always mutually perpendicular.

• Note that if $\vec{E}$  is not perpendicular to a equipotential surface everywhere, it would have a nonzero component along the surface. So to move a test charge against this component work would have to be done. But by the definition of equipotential surfaces, there is no potential difference between any two points on an equipotential surface and hence no work is required to displace the charge on the surface. Therefore, we can conclude that the electrostatic field must be normal to the equipotential surface at every point, and vice versa.

Diagrams :

(a) A pair of charged parallel metal plates sets up a uniform electric field between the plates, away from its edges, as shown by their even spacing. The field is perpendicular to the plates, in the direction from the positive plate toward the negative plate. Conductors are equipotential surfaces, so the negative plate is an equipotential surface with potential taken as zero and the positive plate is an equipotential surface with potential = V. The equipotential surfaces between the plates are parallel to the plates. Electric field lines and equipotential surface.

Advantages of using electrostatic potential :

• The electrostatic potential at each point in space in the vicinity of the source charges represents a scalar field.
• The advantages of electrostatic potential field associated with a given distribution of charges are as follows :
• If we know the potential difference between any two points, we can easily obtain the change in potential energy and the work done when a charge placed in the field moves between these two points.
• Electric field is a vector field. Electrostatic potential being a scalar field, the potential at any point due to several charges is simply the algebraic sum of the potentials due to the individual charges.
• The construction of equipotential surfaces helps to visualize the electric field pattern.
• It is possible to calculate the electric field  from the scalar potential field function V (by differentiating V with respect to the space coordinates.)

(a) Potential energy of a system of 2 point charges:

Let us consider 2 charges q₁ and q₂ with position vectors r₁ and r₂ relative to some origin (O).

To calculate the electric potential energy of the two charge system, we assume that the two charges q₁ and q₂ are initially at infinity.

We then determine the work done in bringing the charges to the given location by an external agency.

In bringing the first charge q₁ to position A($\vec{r_1}$) no work is done because there is no external field against which work needs to be done as charge q₂ is still at infinity i.e., W₁ = 0. This charge produces a potential in space given by

V₁ = $\frac{1}{4πε_0}\frac{q_1}{r_1}$      ……..(1)

Where r₁ is the distance of point A from the origin.

When we bring charge q₂ from infinity to B ($\vec{r_2}$ at a distance r₁₂, from q₁, work done is

W₂ = (potential at B due to charge q₁) × q₂

= $\frac{1}{4πε_0}\frac{q_1}{r_{12}}×q_2$  ...(where AB = r₁₂) ……..(1)

This work done in bringing the two charges to their respective locations is stored as the potential energy of the configuration of two charges.

Hence, the total work done is

W = W₁ + W₂ = 0 +$\frac{1}{4πε_0}\frac{q_1q_2}{r_{12}}$

= $\frac{1}{4πε_0}\frac{q_1q_2}{r_{12}}$

Since the charges were always kept in equilibrium, the change in the potential energy U_f − U_i equals W.

Since the charges were brought from infinity where their potential energy is assumed to be zero, U₁ = 0. Therefore, the potential energy of the system of two point charges is

U = U_f = $\frac{1}{4πε_0}\frac{q_1q_2}{r_{12}}$

(b) Potential energy for a system of N point charges:

Consider assembling a configuration of N point charges q₁, q₂, q₃,... , q_N at points A, B, C, D,…, respectively, in a region free of external electric field.

Let $\vec{r_1}$, $\vec{r_2}$, $\vec{r_3}$, ......$\vec{r_N}$ be the position vectors of the points A, B, C, D,…  etc., respectively, with respect to an arbitrary reference frame.

A configuration of three point charges (For reference only)

No work is done in bringing the first charge q₁ from infinity to point A, so W₁ = 0. Subsequently, the potential at B is

V_B = $\frac{1}{4πε_0}\frac{q_1}{r_{21}}$

where r₂₁ =|$\vec{r}_{21}$|, $\vec{r}_{21}$ = $\vec{r_2}-\vec{r_1}$ being the position vector of B with respect to A. Consequently, the work done by an external agent in bringing q₂ from infinity to B in the electric field of q₁ is

W₂ = V_B.q₂ = $\frac{1}{4πε_0}\frac{q_1q_2}{r_{21}}$

Subsequently, the potential at C is

V_C = V₁ + V₂

where V₁ and V₂ are the potentials due to q₁ and q₂.

V₁ = $\frac{1}{4πε_0}\frac{q_1}{r_{31}}$, V₂ = $\frac{1}{4πε_0}\frac{q_2}{r_{32}}$,

Where r₃₁ =|$\vec{r}_{31}$|, $\vec{r}_{31}$ = $\vec{r_3}-\vec{r_1}$ and r₃₂ =|$\vec{r}_{32}$|, $\vec{r}_{32}$ = $\vec{r_3}-\vec{r_2}$

∴ V_C = $\frac{1}{4πε_0}(\frac{q_1}{r_{31}}+\frac{q_2}{r_{32}})$

Consequently, the work done by the external agent in bringing the third charge q₃ from infinity to C in the electric fields of q₁ and q₂ is

W₃ = V_C.q₃ = $\frac{1}{4πε_0}(\frac{q_1q_3}{r_{31}}+\frac{q_2q_3}{r_{32}})$

Now, the potential at D is

V_D = V₁ + V₂ + V₃ = $\frac{1}{4πε_0}(\frac{q_1}{r_{41}}+\frac{q_2}{r_{42}}+\frac{q_3}{r_{43}})$

And W₄ = V_D.q₄ = $\frac{1}{4πε_0}(\frac{q_1q_4}{r_{41}}+\frac{q_2q_4}{r_{42}}+\frac{q_3q_4}{r_{43}})$

Hence, the total work done in assembling N point charges is

W = W₁ + W₂ + W₃ + W₄+ .... .. +W_N.

= $\frac{1}{4πε_0}[\sum_{j=1\,to\,N-1,J≠k}(\sum_{k=2\,to\,N})\frac{q_jq_k}{r_{kj}}]$

= U_f − U_i

Since the charges were brought from infinity where the potential energy is assumed to be zero, U_i = 0. Therefore, the potential energy of the configuration of N point charges is

U = U_f = $\frac{1}{4πε_0}[\sum_{j=1\,to\,N-1,J≠k}(\sum_{k=2\,to\,N})\frac{q_jq_k}{r_{kj}}]$  = $\frac{1}{4πε_0}[\sum_{\text{all pairs}}\frac{q_jq_k}{r_{kj}}]$

(d) Potential energy of a system of two charges in an external field:

Consider assembling a system of two point charges q₁ and q₂ at points A and B, respectively, in a region of external electric field.

Let $\vec{r_1}$ and $\vec{r_2}$ be the position vectors of A and B, respectively, with respect to an arbitrary reference frame. $\vec{r}_{21}$ =$\vec{r_2}-\vec{r_1}$  is the position vector of B with respect to A.

Let V₁ and V₂ be the electric potentials of A and B due to the external field.

The work done in bringing the charge q₁ from infinity to A against the electric force of the external field is

W₁ = T₁V₁

Subsequently, the electric potential at B is

V_B = V₂ + V’(due to q₁)

= V₂ + $\frac{1}{4πε_0}\frac{q_1}{r_{21}}$

where r₂₁=|$\vec{r}_{21}$|.

Consequently, the work done in bringing the second charge q₂ from infinity to B is

W₂ = q₂V_B

= q₂V₂ + $\frac{1}{4πε_0}\frac{q_1q_2}{r_{21}}$

Hence, the total work done by an external agent in assembling the two point charges in a region of external electric field is

W = W₁ + W₂

= q₁V₁ + q₂V₂ + $\frac{1}{4πε_0}\frac{q_1q_2}{r_{21}}$

Since the charges were always kept in equilibrium, the change in the potential energy of the system U_f − U_i = W. Also, since the charges were brought from infinity where their potential energy is assumed to be zero, U_i = 0. Therefore, the potential energy of the system of two charges in an external field is

U = U_f = q₁V₁ +q₂V₂ + $\frac{1}{4πε_0}\frac{q_1q_2}{r_{21}}$

Reasons of shocks : Metals, humans, earth and animal bodies are all conductors. The main reason we get electric shocks is that being a good conductor our human body allows a resistance free path for the current to flow from the wire to our body.

• During electric shock we experience an extreme stimulation of nerves and muscles. It needs a minimum of 1 mA of electric current to pass through our body for us to experience a shock.
• The danger of electric shock arises not from mere contact with a live wire but rather from simultaneous contact with a live wire and another body of wire at a different potential so that our body provides a conducting path between the two and current passes through our body.
• Touching a single wire by the birds does not result in a current through their bodies because then the electric circuit is not complete.
• But if a person touches two wires at different potentials at once, or if a bare-footed person touches the live wire only, the electric circuit is complete and the person receives an electric shock. In the latter case, the current from the wire passes to the Earth through the body.

Properties of conductors under electrostatic conditions :

Under electrostatic conditions the conductors have following properties.

• In the interior of a conductor, net electrostatic field is zero.
• Potential is constant within and on the surface of a conductor.
• In static situation, the interior of a conductor can have no charge.
• Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point.
• Surface charge density of a conductor could be different at different points.

Electrostatic shielding

The use of a conducting box to protect sensitive instruments from stray electric fields, or the use of a conducting wire cage to protect a person near a high-voltage installation or from lightning strike, is called electrostatic shielding

• When an isolated conductor, uncharged or charged, is placed in an external electric field, as in Fig. all points of the conductor come to the same potential. The free conduction electrons in the conductor distribute themselves on the surface, leaving a net positive charge on some regions of the surface and a net negative charge on other regions.

• This charge distribution causes an additional electric field at interior points such that the total field at every point inside is zero.
• The charge distribution on the conductor is such that the net electric field at all points on the surface to be perpendicular to the surface, thereby altering the shapes of the field lines near the conductor.
• During lightning and thunder storm it is always advisable to stay inside the car than near a tree in open ground, since the car acts as a shield.

Faraday cage :

The hollow conductor or the conducting wire cage that shields its interior from external electric fields is called a Faraday cage or Faraday shield

• A Faraday Cage, made from a contiguous metal sheet or from a fine metal mesh, is used to shield its content or occupant from static and nonstatic electric fields.
• Electro-magnetic shielding: MRI scanning rooms are built in such a manner that they prevent the mixing of the external radio frequency signals with the MRI machine.

Polarization of a non-polar dielectric in an external electric field:

In the absence of an external electric field, the molecules of a nonpolar dielectric have no inherent electric dipole moments. An applied electric field slightly separates the centres of negative and positive charges. Then, a nonpolar molecule acquires an induced dipole moment in the direction of the applied field, as shown in Fig.

The induced dipole moments of all the molecules add up giving the dielectric a net induced electric dipole moment in the presence of the applied field.

Polarization of a polar dielectric in an external electric field:

In the absence of an external electric field, the permanent electric dipole moments of the molecules of a polar dielectric orient in random directions due to thermal agitation such that their vector sum is zero, see below Fig. (a).

An applied electric field does slightly increases the separation between the centres of negative and positive charges. But, a much larger effect is the tendency of the dipole moments to align with the field, although thermal agitation prevents complete alignment, as shown in above Fig. (b).

• Due to the partial alignment of the dipole moments, a polar dielectric also acquires a net induced electric dipole moment in the direction of the applied field.
• The extent of polarisation depends on the relative values of the two opposing tendencies: (1) the tendency of the applied field to align the dipoles (2) thermal agitation that tends to randomize.

Reduction of electric field due to polarization of a dielectric:

When a dielectric is placed in an external electric field, the value of the field inside the dielectric is less than the external field as a result of polarization.

Consider a rectangular slab of a linear isotropic dielectric placed in a uniform external electric field. In case of a nonpolar dielectric, the applied field slightly separates the centres of negative and positive charge in a molecule inducing an electric dipole moment in the direction of the field. In most cases, this separation is a very small fraction of a molecular diameter. In case of a polar dielectric, the permanent dipole moments of its molecules are partially aligned with the field. In either case, the dielectric is said to become polarized.

In a uniformly polarized dielectric, the charges of adjacent interior dipoles add to zero. But due to the unbalanced positive ends of dipoles at one face of the slab, bound positive charge appears on that exterior surface. Similarly, bound negative charge appears on the opposite exterior surface of the slab. These bound surface charges are called polarization charges. There is, however, no excess charge in any volume element within the slab and the slab as a whole remains electrically neutral.

The external electric field $\vec{E_0}$ polarizes the dielectric, with a net polarization $\vec{P}$ parallel to $\vec{E_0}$. Within the dielectric, the induced field $\vec{E_P}$ = $-\frac{\vec{P}}{ε_0}$ due to the polarization charges is opposite to the applied field, see Fig.

$\vec{E}$ = $\vec{E_0}$+$\vec{E_P}$ = $\vec{E_0}$$-\frac{\vec{P}}{ε_0}$

i.e. E = E₀ − E_P in magnitude

Assuming the dielectric to be isotropic and linear,

$\vec{P}$= χ_eε₀$\vec{E}$

where χ_e is the electric susceptibility of the dielectric.

∴ $\vec{E_0}$ = (1 + χ_e)$\vec{E}$  = k$\vec{E}$

where k = 1 + χ_e is the dielectric constant. This implies E = E₀/k, thus less than E₀. Thus, the effect of a dielectric material is always to decrease the electric field below the applied electric field.