A capacitor is a device that stores the electric charge as the potential difference between the two plates and the electric field in the capacitor is on the application of the voltage source.

**The potential difference is created by the transportation of electrons from the positive terminal to the negative terminal of the capacitor and establishing the electric field in a capacitor. This charge difference stores the electric energy in the form of the potential of the charge and is proportional to the charge density on each plate.**

**Electric Field in Capacitor Formula**

Like positive and negative charges, the capacitor plate also behaves as an acceptor and donor plate when the source is passed through the capacitor plates. The positive terminal of the capacitor will donate the electron and these free electrons will be accepted by the negative terminal of the capacitor.

**Due to the mobility of the free charges, the electric flux will be introduced within the capacitor and the total electric field in the capacitor will be**

E=δ/∈_{0}

**The charge density of each capacitor plate is called the surface density which is stated as the charge present on the surface of the plate per unit area and is given as σ** **=Q/A.**

Hence,

This equation gives the electric field produced between the two plates of the capacitor.

**Electric Field Inside a Capacitor**

**The capacitor has two plates having two different charge densities. The electric flux passes through both the surfaces of each plate hence the Area = 2A.**

Consider two plates having a positive surface charge density and a negative surface charge density separated by distance ‘d’. Let A be the area of the plates. The electric flux line is running from the positively charged plate to the plate with a majority of the negative carriers as depicted in the below figure.

Let P be any point in the middle of the two charged plates of the capacitor.

By applying Gauss law,

φ =EA

The electric field due to one charged plate of the capacitor is

E.2A= q/ε_{0}

We know that σ =Q/A

Using this in the above equation

**Hence, the resultant electric field at any point between the plates of the capacitor will add up.**

Inserting value for σ, we get

This is the total electric field inside a capacitor due to two parallel plates.

**What is the electric field produced by the parallel plate capacitor having a surface area of 0.3m**^{2} and carrying a charge of 1.8C?

^{2}and carrying a charge of 1.8C?

**Given:** q=1.8C

A= 0.3m^{2}

We have

=0.68 x 10^{12} V/m

The electric field produced by the parallel capacitor carrying a charge of 1.8 C is 0.68 x 10^{12} V/m.

**Electric Field Outside a Capacitor**

Now, if point P lies outside the capacitor then the electric field at point P due to the plate having a positively charged surface density is

Whereas, the electric field at point P due to negative charge surface density plate of the capacitor is

Hence, the net electric field due to both the plates of the capacitor is

E=E_{1}+E_{2}

E=0

**At any point outside the capacitor, the electric field is always zero. Because, on supplying the electric current through the capacitor, one terminal of the capacitor will have a positive surface charge density and another will have a negative surface charge density.**

**Electric Field in Capacitor With Dielectric**

Now we know that in presence of vacuum, the electric field inside a capacitor is E=σ/ε_{0} , the potential difference between the two plates is V=Ed where d is a distance of separation of two plates and hence the capacitance in this case is

C= Q/V = ε_{0}A/d

Now if we place dielectric between the two plates of the capacitor on polarization, occupying the complete space between the two plates, the surface charge densities of the two plates are +σ_{p} and –σ_{n}. Thus the net surface charge density of both the plates is

Hence, the electric field through the capacitor is

Thus, the potential difference becomes

For linear dielectrics,

So,

Where k is a dielectric constant and is greater than 1 i.e. k>1.

Hence, the potential difference now becomes

Inserting value for surface charge density

V= Qd/Aε_{0}k

Hence, the capacitance of the capacitor is

C= Q/V = ε_{0}kA/d

ε_{0}k is a permittivity of medium and is denoted as ε

Therefore the equation now becomes

C= εA/d

**What is the electric field and the potential difference of a capacitor in presence of a dielectric medium of permeability 6× 10**^{-12} C^{2}N^{-1}m^{-2} of width 3cm if surface charge densities are 6 C/m^{2} and -5.8 C/m^{2}?

^{-12}C

^{2}N

^{-1}m

^{-2}of width 3cm if surface charge densities are 6 C/m

^{2}and -5.8 C/m

^{2}?

**Given:** σ_{p}=6 C/m^{2}

σ_{n}=-5.8 C/m^{2}

ε_{0}= 6 x 10^{-12} C^{2}N^{-1}m^{-2}

d=3cm=0.03m

The electric field of the capacitor is

The electric field of the capacitor is found to be 3.3 x 10^{10} V/m, thus the potential difference between the capacitor plates is

V=Ed

=3.3 x 10^{10} x 0.03

=0.099 x 10^{10} V

=0.1 x 10^{10} V

The potential difference between the two capacitor plates is 0.1 x 10^{10 }V.

**Electric Field Capacitor in Series**

When capacitors are connected in series, the potential difference between the plates adds up. If we have two capacitors C_{1} and C_{2} connected in series, and the potential difference across the plates is V_{1} and V_{2} respectively, then the net potential difference becomes

V=V_{1}+V_{2}

The capacitance is C= Q/V

Hence, V=Q/C

Using this in the above equation we get

V=Q/C_{1} + Q/C_{2}

Solving this further

The potential difference is also equal to V=Ed

Hence the electric field due to capacitors in series we can calculate as

E= V/d

If there are ‘n’ numbers of capacitors connected in series then the electric field across the n capacitors will be

**Electric Field in Cylindrical Capacitor**

**A cylindrical capacitor consists of two cylindrical plates. The inner cylinder has a positive surface charge density +σ of radius ‘r’ and the outer cylinder has a negative surface charge density –σ having a radius ‘R’.**

The electric flux runs from the surface of the inner cylinder to the outer cylinder as shown in the above figure. Fig (b) shows the cross-sectional view of the cylindrical capacitor. Let ds be the Gaussian surface at the middle of the two charged cylinders.

The electric field inside the inner cylinder is zero as there is no electric flux through this region and as well as outside the cylinder of radius ‘R’ is also zero. The electric flux is running between the two cylinders at a distance s from the center.

The electric flux through the Gaussian surface ds is given by

Therefore,

This equation gives the electric field produced by the cylindrical capacitor.

**What is the electric field at a point 0.6 cm away from the center of a cylindrical capacitor of the height of 2cm having an outer radius of 0.8 cm and the inner radius of 0.35 cm carrying a charge of 5C?**

**Given:** r= 0.35cm= 0.0035m

R= 0.8cm= 0.08m

S= 0.6cm= 0.06m

h=2cm= 0.02m

Q=5C

We have,

=74.62 x 10^{12} V/m

The electric field of the capacitor at a distance of 0.6cm from the center of the cylindrical capacitor is 74.62 x 10^{12} V/m.

**Electric Field Intensity in Capacitor**

The electric field intensity outside the charged capacitor region is always zero as the charge carriers are present on the surface of the capacitor.

**In the inner region of the capacitor, the electric field is equal to the ratio of the density of the surface charge carriers, and the permeability of the medium in this region is the same at all the points inside the capacitor.**

Where σ is the surface charge density of the charge carriers present on the plate of the capacitor and

ε_{0 }is the permeability of the medium

Also, the electric field can be calculated by measuring the potential difference between the two plates and finding the distance of separation of plates as

E=V/d

Where V is a potential difference between the plates of the capacitor and

d is the distance between the two plates

**Electric Field in Spherical Capacitor**

Like a cylindrical capacitor, the spherical capacitor also consists of two spheres having oppositely charge carriers on the surfaces of each sphere.

Consider a sphere of radius ‘R_{2}’ having a surface charge density as +σ and another sphere of radius ‘R_{1}’ of surface charge density –σ covering the small spherical shell.

**The electric flux runs from the sphere consisting of a positive surface charge density to the outer sphere.** Consider a Gaussian surface ‘ds’ in the middle of the two spherical surfaces at a distance ‘r’ from the center of the spheres. Let the charge be q at the Gaussian surface. By applying gauss law

S is a surface area that is equal to 4πr^{2}, hence we get

E x 4πr^{2} = q/ε_{0}

The electric field in the spherical capacitor is

E= q/4πε_{0}r^{2}

The potential difference between the two charged spheres is

We have found out the electric field of the spherical capacitor, thus let us substitute the same in this equation.

Hence the capacitance of the spherical capacitor is

C= q/V

Inserting the value of the potential difference, we get

This equation gives the capacitance of the spherical capacitor.

**Frequently Asked Questions**

**What is the electric field of a charged sphere of radius 3cm carrying a charge of 4C?**

**Given:** r=3cm=0.03m

Q=4C

The electric field inside the sphere is E=0.

The surface area of the sphere is

A=4πr^{2}

=4π x (0.03)^{2}

=0.01 m^{2}

Hence, the surface charge density of a sphere is

σ = Q/A

= 4C/0.01m^{2}

=400 C/m^{2}

Therefore the electric field of a charged sphere is

=45.2 x 10^{12} V/m

The electric field at the surface and at a point outside the sphere is 45.2 x 10^{12} V/m.

**What is the electric field of the spherical capacitor at a distance of 4cm from the center having an inner radius of 3cm and an outer sphere of 5cm carrying a charge of 2mC?**

**Given:** R_{1}=3cm=0.03m

R_{2}=5cm=0.05m

r= 4cm= 0.04m

q= 2mC

The electric field at a Gaussian surface at a distance of 0.04m from the center of the spherical capacitor is

=11.23 x 10^{6} V/m

The capacitance is the spherical capacitor is

=8.3 x 10^{-12}F