## Grade 11 Physics Note

# Electrostatic force, Field and potential

**Relative permittivity or dielectric constant of mediums:**

The **dielectric constant** - also called the **relative permittivity** **-** indicates how easily a material can become polarized by imposition of an electric field on an insulator. Relative permittivity - ε - is the ratio of the permittivity of a substance to the permittivity of free space or vacuum.

**Electron volt (eV);**

The energy gained by an electron which has been accelerated through a potential difference 0of one volt is called an electron volt (eV).

**A charged particle is free to move in an electric field;**

If the charged particle was initially at rest it will move along an electric line of force. If the initial velocities of the charge particles make a certain angle with a line of force, then the charge particles shall not move along the line of force.

**Gauss’s theorem;**

This theorem states that the total electric flux through any closed surface surrounding a charge is equal to the net positive charge enclosed by that surface.

A Gaussian surface is a closed surface in three-dimensional space through which the flux of a vector field is calculated; usually the gravitational field, the electric field, or magnetic field.

**Equi – potential surfaces:**

The work done to move a charge from any point on the equi-potential surface to any other point on the equi-potential surface is zero since they are at the same potential. Further, equi-potential surfaces are always perpendicular to the net electric field lines passing through it.

The work done to move a charge from any point on the equi-potential surface to any other point on the equi-potential surface is zero since they are at the same potential. Further, equi-potential surfaces are always perpendicular to the net electric field lines passing through it.When two surfaces intersect then the value of potential on both the surfaces never be same due to electro static induction that's why the equi-potential surfaces cannot intersect each other.

**Electric potential and potential gradient:**

An electric potential (also called the electric field potential or the electrostatic potential) is the amount of electric potential energy that a unitary point electric charge would have if located at any point in space, and is equal to the work done by an electric field in carrying a unit positive charge from infinity.

A potential gradient is the local rate of change of the potential with respect to displacement, i.e. spatial derivative, or gradient.

**Electric intensity at a point in an electric field;**

*Electric potential* at a *point* charge at a *point* is *defined* as the amount of work done in bringing a unit positive test charge from infinity to that *point *against the *electric field* of the *point* charge. It is scalar quantity.

**Columbs law in electrostatics:**

Coloumb's Law is applied to calculate the force of attraction or repulsion between two-point charges.

It states that "the force of attraction or repulsion between two charges is directly proportional to the product of charges and inversely proportional to the square of the distance between them."

Figure: Two charges separated by a distance r apart.

Suppose two charges Q_{1} and Q_{2} separated at a distance r as shown in the figure.

According to Coloumb's Law, the force, F between the charges is

1. Directly proportional to Q_{1} Q_{2}

2. Inversely proportional to r^{2}

${\rm{F}} \propto \frac{1}{{{{\rm{r}}^2}}}{\rm{\: }}$……………………………..1

${\rm{F}} \propto \frac{{{\rm{Q}}1{\rm{Q}}2}}{{{{\rm{r}}^2}}}$……………………………2

Combining both we get,

${\rm{F}} = {\rm{k}}.\frac{{{\rm{Q}}1{\rm{Q}}2}}{{{{\rm{r}}^2}}}$

Where, k is constant of proportionality. Its value depends upon the medium between the charges and the system of unit used.

**Gauss’s theorem in electrostatics;**

This theorem states that the total electric flux through any closed surface surrounding a charge is equal to the net positive charge enclosed by that surface.

For conductor, the charges lie only on the surface. Therefore, in case of a hollow sphere also, the charge lie only on the surface. And hence the expressions for electric field intensity at various points are similar to those for the solid charged sphere.

a. For a point p outside the hollow charged sphere, the electric field is

${\rm{E}} = \frac{{\rm{q}}}{{4{\rm{\pi }}{{\rm{E}}_0}{{\rm{r}}^2}}}$,r>R.

b. For a point p on the surface of the hollow charged sphere, r=R

E=$\frac{{\rm{q}}}{{4{\rm{\pi }}{{\rm{E}}_0}{{\rm{R}}^2}}}$,r>R.

For point p inside the hollow charged sphere, the charge enclosed by the Gaussian surface will be zero. Therefore, the electric field intensity E at the point inside the hollow charged sphere will be E=0

Gauss’s theorem:

This theorem states that the total electric flux through any closed surface surrounding a charge is equal to the net positive charge enclosed by that surface.

The electric field intensity E at point p is perpendicular to the surface area A but parallel to the curved surface of the cylinder.

So, the flux passing normally through the surface area A is given by

ϕ +E * A

The net charge Q enclosed by the Gaussian surface is

Q = σ * A

From Gauss’s theorem, the total flux passing through the cylinder is

ϕ = $\frac{{\rm{Q}}}{{{{\rm{E}}_0}}}$

Or, ${\rm{E}} = \frac{\sigma }{{{{\rm{E}}_0}}}$

(i) Outside the sphere;

Total flux passing through the Gaussian surface is given by

ϕ = E * A = E * 4πr^{2}

The charge enclosed by the Gaussian surface is q. thus, if the charged sphere is in Vacuum, applying Gauss’s theorem we will have

Flux = $\frac{{{\rm{charge\: enclosed}}}}{{{{\rm{E}}_0}}}{\rm{\: }}$

ϕ = $\frac{{\rm{Q}}}{{{{\rm{E}}_0}}}$

E=$\frac{{\rm{q}}}{{4{\rm{\pi }}{{\rm{E}}_0}{{\rm{r}}^2}}}$

(ii) Inside the sphere

Gaussian surface is given by

ϕ = E * 4πr^{2}

The charge enclosed by Gaussian surface thus constructed is Zero.

From the Gaussian theorem,

ϕ = $\frac{{\rm{q}}}{{{E_0}}} = 0$

E * 4π r^{2}=0

E = 0.

Hence there is no electric field inside a solid charge sphere.

(iii) At the surface of sphere.

Flux through the Gaussian surface is given by

ϕ = E * 4π ^{2 }= E * 4π R^{2}

And ϕ = $\frac{{\rm{Q}}}{{{E_0}}}$

${\rm{E}} = \frac{{\rm{q}}}{{4{\rm{\pi }}{{\rm{E}}_0}{{\rm{R}}^2}}}$

**Electric lines of force;**

In order to point out the direction of an electric field we can draw a number of lines called electric lines of force. Lines of force originate from a positive charge and terminate to a negative charge. The tangent to the line of force indicates the direction of the electric field and electric force.

Properties:

1. The electric lines of force are imaginary lines.

2. The electric lines of force emanate from a positive charge and terminate on a negative charge.

3. The tangent to an electric field line at any point gives the direction of the electric field at that point.

4. Two electric lines of force can never cross each other. If they do, then at the point of intersection, there will be two tangents. It means there are two values of the electric field at that point, which is not possible. Further, electric field being a vector quantity, there can be only one resultant field at the given point, represented by one tangent at the given point for the given line of force.

5. Electric lines of force are closer (crowded) where the electric field is stronger and the lines spread out where the electric field is weaker.

6. Electric lines of force are perpendicular to the surface of a positively or negatively charged body.

7. Electric lines of force contract lengthwise to represent attraction between two unlike charges.

8. Electric lines of force exert lateral (sideways) pressure to represent repulsion between two like charges.

9. Electric lines of force can pass through an insulator.

**Expression for the potential at a point due to a point charge:**

An electric potential (also called the electric field potential or the electrostatic potential) is the amount of electric potential energy that a unitary point electric charge would have if located at any point in space, and is equal to the work done by an electric field in carrying a unit positive charge from infinity.

The electric potential created by a point charge Q, at a distance r from the charge (relative to the potential at infinity), can be shown to be

${{\rm{V}}_{\rm{E}}} = \frac{1}{{4{\rm{\pi }}{{\rm{E}}_0}}}{\rm{\: }}\frac{{\rm{Q}}}{{\rm{r}}},$

Where, ε_{0} is the electric constant (permittivity of vacuum). This is known as the Coulomb potential.

The electric potential due to a system of point charges is equal to the sum of the point charges' individual potentials. This fact simplifies calculations significantly, since addition of potential (scalar) fields is much easier than addition of the electric (vector) fields.

**An expression for the potential difference between two point r _{1} and r_{2} from +Q;**

An electric potential (also called the electric field potential or the electrostatic potential) is the amount of electric potential energy that a unitary point electric charge would have if located at any point in space, and is equal to the work done by an electric field in carrying a unit positive charge from infinity.

Consider a point charge +Q placed at point O in free space as shown in free space. It is desired to find electric potential at point A at a distance r from O, i.e. OA = r

If W ∞_{A} is the amount of work done, the potential at point is given by

V_{A }= W∞ A……………………1

To calculate W ∞_{A}, consider that at any instant, the test charge is at point P which is at a distance X from O such that OP = X.

F = $\frac{{{\rm{Q*}}1}}{{4{\rm{\pi }}{{\rm{E}}_0}{{\rm{X}}^2}}}{\rm{\: }}$= $\frac{{\rm{Q}}}{{4{\rm{\pi }}{{\rm{E}}_0}{{\rm{X}}^2}}}$………………..2

If the test charge is moved dx from P to Q

dW = -Fdx$\frac{{\rm{Q}}}{{4{\rm{\pi }}{{\rm{E}}_0}{{\rm{X}}^2}}}{\rm{dx}}$……………….3

The total amount of work done in bringing the test charge from infinity to A is given by integrating Eq 3, so,

W ∞_{A}=$\mathop \smallint \limits_{\rm{\infty }}^{\rm{A}} {\rm{dW}} = \mathop \smallint \limits_{\rm{\infty }}^{\rm{r}} - {\rm{\: }}\frac{{\rm{Q}}}{{4{\rm{\pi }}{{\rm{E}}_0}{{\rm{X}}^2}}}{\rm{dx}}$

$\frac{{\rm{Q}}}{{4{\rm{\pi }}{E_0}}}\left[ {\frac{1}{{\rm{r}}} - \frac{1}{{\rm{\infty }}}} \right] = \frac{{\rm{Q}}}{{4{\rm{\pi }}{E_0}}}\left[ {\frac{1}{{\rm{r}}} - 0} \right]$

W ∞_{A}= $\frac{{\rm{Q}}}{{4{\rm{\pi }}{{\rm{E}}_{0{\rm{\: r}}}}}}$…………4

From 1 and 4

${{\rm{V}}_{\rm{A}}} = \frac{{\rm{Q}}}{{4{\rm{\pi }}{{\rm{E}}_0}{\rm{r}}}}$

This is the electric potential at point A which is at a distance r from the point charge +Q.

**A relation between the potential gradient and electric field strength;**

The change is electric potential in the space around a charge with respect to the change in distance is called potential gradient.

An electric field potential is the amount of electric potential energy that a unitary point electric charge would have if located at any point in space, and is equal to the work done by an electric field in carrying a unit positive charge from infinity.

A potential gradient is the local rate of change of the potential with respect to displacement, i.e. spatial derivative, or gradient.

Potential gradient is defined as g = dV / dx

Electric field is defined as E = - dV / dx

So, E = - g

Thus, the potential gradient g at a point in the electric field is negative in magnitude electric field E at that point.

**Equipotentials and line force around a point charge****:**

The work done to move a charge from any point on the equi-potential surface to any other point on the equi-potential surface is zero since they are at the same potential. Further, equi-potential surfaces are always perpendicular to the net electric field lines passing through it.

Figure: Equipotentials and line force around a point charge

The electric potential of apoint charge is given by

${\rm{V\: }} = \frac{{{\rm{kQ}}}}{{\rm{r}}} = \frac{{\rm{Q}}}{{4{\rm{\pi }}{{\rm{E}}_0}{\rm{r}}}}$

So that the radius r determines the potential,the equipotential lines are therefore circles and a sphere centered on the charge is an equipotential surface. The dashed lines illustrate the scaling of voltage at equal increments - the equipotential lines get further apart with increasing r.

**If an electron is released from the rest in an electric field, it starts to move from the point of lower potential to higher potential;**

As a general rule in electrostatics, electric field lines point in the direction of decreasing electric potential. If a negative charge is initially at rest in an electric field, it will tend to move opposite the electric field. Therefore, a negative charge naturally moves in the direction of increasing potential. A positive charge, on the other hand, tends to move with the electric field so, as a result, it naturally moves in the direction of decreasing potential.