What is the effect on the electric field when increasing the distance from the charge 2 times?

A capacitor is meant to store charges. The ability to store how much charges is measured as the device's capacitance. It is related to the voltage across the capacitor plates and the charge on the plates:

C=Q/V

Consider a parallel plate capacitor as shown:

Suppose initially the charge on the plate is zero. Now you connect it to an emf. The charges get accumulate on the plate in the same polarity as the emf. It takes some time for the plate to get charged to maximum charge Q. This charge on the plate induce a negative charge of same amount on the other plate. In between the plates, we have a dielectric and let the separation distance of the plated be d.

Now as we increase the area of the plates, more charges can occupy the plates and hence increase it's ability to store charges, or increase it's capacitance. Since we have some charges on a surface, each plate develops electric field and this electric filed is equal and opposite for the two charged plates. A glance on the direction of these fields indicate that the net electric field exists only in between the plates while outside the plates, the net field vanishes. The net electric field inside the plate will be equal to twice that due to the individual plate, given by:

E=σ/$ϵ_0 $

where σ- surface charge density (Q/A)
$ϵ_0 $- permittivity of free space. In a medium of dielectric constant $ϵ_r$, the $ϵ_0$ has to be replaced by ϵ=$ϵ_r$$ϵ_0$

This electric field is constant throughout the space between the plates as you can see the field does not depends on the distance. Now the electric potential developed between the plates is nothing but the gradient of the electric field.

E=V/d

So we write

C=QE/d or

C=$ϵ_0$A/d

which means that the capacitance of a plate is dependent on the distance between the plates.

Or there is a simple way to remember this:

On increasing the area of the plates, you could accommodate more charges on the plates and this in turn will increase the electric field between the plates. Increase in electric field between the plates means the voltage across the plates increase as E=V/d. Also the p.d between the plates increases with decrease in d. Hence we write, the capacitance as:

C∝$A/d$

The proportionality constant is the permittivity of the medium ϵ.

So, C=$ϵA/d$

A decrease in d corresponds to decrease in V (V=Ed) and increases the capacitance C.

Of course increase in C increases the time constant T=RC, if the discharge occurs through a resistor.

• electric field between parallel charged plates `E=V/d`

• acceleration of charged particles by the electric field `F_Net=ma, F=qE`

• work done on the charge `W=qV`, `W=qEd`, `K=1/2mv^2`

Electric Field Strength

• Two parallel charged plates connected to a potential difference produce a uniform electric field of strength:

`E=V/d`

where E is the electric field strength (Vm-1 or NC-1), V is the potential difference (voltage) in volts (V), and d is the perpendicular distance between the two parallel plates in metres (m).

• Note: ‘d’ is strictly the distance between two charged plates, it is not dependent on the position of a particle within the electric field.

• The direction of such an electric field always goes from the positively charged plate to the negatively charged plate (shown below).

• Example: In the electric field above, the electric field strength is:

`E=10/5=2` Vm-1 downwards 

• A charged particle experiences a force when in an electric field.
  • Positively charged particles are attracted to the negative plate
  • Negatively charged particles are attracted to the positive plate

• The magnitude of this force is given by the equation:

`F_E=qE`
 

Where F is the force (N), q is the charge of the particle (C), and E is the electric field strength

• Direction of force depends on the nature of particle’s charge.
  • The charge of the particle is either given by the question or provided in the reference sheet

• The electric field strength can therefore be also expressed in the form:

`E=F/q`

Since:

Therefore:

`F/q=V/d` 

• By Newton’s second law (F=ma), any charged particle in an electric field experiences acceleration. Hence, their change in displacement increases with time (path of motion is curved not linear).

Work Done in an Electric Field

• Since the force acting on a charged particle can be determined by its charge (C), electric field strength (E), potential difference between charged plates (V) and distance between them (d), work done is expressed as such:

`W=Fs`

Substitute force with qE and displacement s with distance d between a pair of charged metal plates:

Alternatively, d can also be expressed in terms electric field strength and potential difference across the metal plates:

`d=V/E`

Thus,

where d is displacement and only equals to distance between two charged plates if the charged particle is moved from one plate to another. In other words, the second equation only applies when work is done to move a charged particle against the electric field. (diagram below) 

• Work done by electric field can be analysed by a change in kinetic energy of the charged particle 

`W=DeltaK = 1/2m(v-u)^2`

Where u and v are initial and final velocities of the charged particle respectively.

Practice Question 1

An electron travelling at 50 m/s enters a uniform electric field created by a pair of parallel metal plates connected to a potential difference of 7 V.

(a) Calculate the electric field strength. (1 mark)

(b) Calculate the acceleration experienced by the electron upon entering the electric field. (2 marks)

(c) After a short while, the electron is found stationary on the surface of the positive plate. Calculate the work done to move the electron from the positive plate to the negative plate. (1 mark)

Qualitative Comparison

Quantitative Comparison

• Charged particles experience very little and negligible amount of gravitational force.
• For example, for an electron on the surface of Earth it experiences gravitational force of magnitude:

`F_g=((6.67xx10^-11)(6.0xx10^24)(9.109xx10^-31))/(6371xx10^3)^2`

`F=9.0xx10^-30` N towards the centre of Earth

• Compared with typical electric fields, the contribution from electric force is much more significant than gravitational force.

Next section: Charged Particles in Magnetic Fields 

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