Contents
Physics Topics cover a broad range of concepts that are essential to understanding the natural world.
Electric Field Intensity : Formulas, Properties and Solved Examples
Electric field: The space around any electric charge in which its influence can be felt, so that any other charge in that region is acted upon by a force, is known as the electric field
of that charge.
Theoretically, the field extends up to infinity, but limited sensitivity of detecting instruments predicts that a charge exerts its influence (force of attraction or repulsion) only over a limited region.
Intensity of on electric field: The intensity or strength of an electric field at any point in the field Is the force experienced by a unit positive charge placed at that point.
Obviously, intensity of an electric field is a vector quantity. The direction of intensity is given by the direction of force acting on the positive charge. An electric field may be generated due to more than one charge. Intensity at a point in such an electric field is determined by calculating the field at that point due to all the charges individually and then the resultant of all the field strengths are found out by rcctor addition. The intensity or
strength of the electric field at a point is generally denoted by the letter \(\vec{E}\). Often it is simply called an ‘electric field \(\vec{E}\)‘.
Suppose, a test charge q0 experiences a force of \(\vec{F}\) when it is placed at a point in an electric field. Then, the electric field at that point is \(\vec{E}\) = \(\frac{\vec{F}}{q_0}\).
Here, the test charge q0 should be very small so that it cannt alter the original electric field. So the correct definition of E should be,
\(\vec{E}\) = \(\lim _{q_0 \rightarrow 0} \frac{\vec{F}}{q_0}\)
1. There is no problem as such in using the definition of \(\vec{E}\), except that we cannot make q0 arbitrarily small, because in the real world, we have never observed a charge smaller than the electronic charge.
2. No problem arises however, if we take the definition of \(\vec{E}\) of a charge distribution at any point (x, y, z), where the position vector is \(\vec{r}\)(= \(\hat{i}\)x + \(\hat{j}\)y + \(\hat{k}\)z) as
\(\vec{E}\) = \(\frac{1}{4 \pi \epsilon_0} \sum_{j=1}^n \frac{q_j \hat{r}_j}{r_j^2}\)
where \(\hat{r}_j\) is the unit vector from the jth charge to the point(\(\vec{r}\)) (x, y, z).
Unit: The unit of electric field intensity in CGS system is dyn/esu of charge or dyn/statcoulomb and in SI the unit is newton/coulomb (N ᐧ C-1).
An alternative unit of electric intensity in SI :
\(\frac{\mathrm{N}}{\mathrm{C}}\) = \(\frac{\mathrm{N} \times \mathrm{m}}{\mathrm{C} \times \mathrm{m}}\) = \(\frac{\mathrm{J}}{\mathrm{C} \times \mathrm{m}}\) = \(\frac{\mathrm{V}}{\mathrm{m}}\) = V ᐧ m-1
Dimension:
Relation between CGS and SI units of intensity:
1N ᐧ C-1 = \(\frac{10^5 \text { dyn }}{3 \times 10^9 \text { esu of charge }}\) = \(\frac{1}{3 \times 10^4}\) dyn. statC-1
So, 1 esu electric intensity = 3 × 104 N ᐧ C-1
Electric field due to a point charge: Let a point charge q be placed at a point in a medium of permittivity k. This charge will Create an electric field around it. To determine the electric field at a point at a distance r from the charge, a unit positive charge is imagined to be placed at that point. The force experienced by the unit positive charge is the magnitude of electric field at that point.
If a charge q be placed at a point in a medium of permittivity ε, the electric field at a distance r from the charge is given by,
E = \(\frac{1}{4 \pi \epsilon} \cdot \frac{q}{r^2}\) = \(\frac{1}{4 \pi \kappa \epsilon_0} \cdot \frac{q}{r^2}\) ….. (1)
[k = dielectric constant of the medium]
For vacuum or air, k = 1
∴ E = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{r^2}\) ….. (2)
Therefore, an electric field due to a point charge is directly pro portional to the charge (q) and inversely proportional to the square of the distance (r) from it. It is obvious that electric field is different at different points of a field. Since we have assumed for definition that the test charge is positive, the electric field due to this will be directed away from the charge. In case of the source charge is negative, the field will he directed towards the charge.
If E be the electric field at a point, the force acting on a charge q placed at that point, by definition is given by, F = qE
ie.. force = charge × field Intensity
The force acting on a charge in an electric field does not depend on its mass.
In CGS system, for air or vacuum,
F = \(\frac{q}{r^2}\)
It should be noted that If the magnitude and direction of an electric field is the same at all points in the field, it is called a uniform electric field.
Numerical Examples
Example 1.
A negative charge of 20 unit Is placed at a distance 50 cm away from a positive charge of 80 unit. Where will the electric field be zero on the line joining the two charges?
Solution:
Suppose. + 80 unit and -20 unit of charges are placed at A and B respectively [Fig.]. The point where the electric field will be zero cannot lie in between A and B, because in that
case intensity could he along the same direction, i.e., along AB for both the charges.
As the charge at A is greater, the point where the resultant intensity is zero will be situated on the right side of B, say at P.
Suppose, BP = x
Electric field at P due to the charge at B = \(\frac{20}{x^2}\); along \(\overrightarrow{P B}\)
Electric field at P due to the charge at A = \(\frac{80}{(50+x)^2}\); along \(\overrightarrow{A P}\)
Since, the resultant intensity at P = 0
∴ \(\frac{20}{x^2}\) = \(\frac{80}{(50+x)^2}\) or, x = 50, –\(\frac{50}{3}\)
Now, x ≠ –\(\frac{50}{3}\) cm, because the point in that case would be situated in between A and B.
So, x = 50 cm; the point where the field is zero at a distance of 50 cm from the -20 unit charge on its right side.
Example 2.
At each of the four vertices of a square of side 10 cm there is a +20 esu of charge. Find the Intensity of electric field at the point of intersection of the two diagonals. [HS ‘05]
Solution:
The point of intersection O of the diagonals is equidistant from four equal charges [Fig.]. So the electric field at O due to each charge is the same in magnitude (= E, say).
Now, the fields E along \(\overrightarrow{O C}\) and E along \(\overrightarrow{O A}\) cancel each other.
Similarly, E along \(\overrightarrow{O D}\) and E along \(\overrightarrow{O B}\) also cancel.
Therefore, intensity of the electric field at the point of intersection of the two diagonals of the square is zero.
Example 3.
AB and CD are two perpendicular diameters of a circle of circumference 20π cm. There are + 10 esu, +10 esu, +10 esu and -10 esu of charges at A, B, C and D respectively. What is the intensity of electric field at the centre O of the circle? What is the direction of the field?
Solution:
Suppose, O is the centre of the circle (Fig.).
Radius of the circle = r.
Here, 2πr = 20π or, r = 10 cm
Intensities of electric field at O due to the charges at A and B cancel each other, because the two fields are equal and opposite.
Intensity at O duc to the charge at C
= \(\frac{10}{(10)^2}\) = \(\frac{1}{10}\) dyn ᐧ statC-1 ; along \(\overrightarrow{O D}\)
Intensity at O due to the charge at D
= \(\frac{10}{(10)^2}\) = \(\frac{1}{10}\) dyn ᐧ statC-1 ; along \(\overrightarrow{O D}\)
∴ Resultant intensity at O
= (\(\frac{1}{10}\) + \(\frac{1}{10}\)) = 0.2 dyn ᐧ statC-1 ; along \(\overrightarrow{O D}\)
Example 4.
The bob of a pendulum of weight 80 mg carries a charge of 2 × 10-8 C. The bob is at rest in a horizontal electric field of magnitude 2 × 104 V ᐧ m-1. Determine the tension in the string and angle of the string with the
vertical. Given, g = 9.8 m s-2.
Solution:
Suppose, the string of the pendulum is inclined at an angle of θ with the vertical [Fig.]
Let the horizontal electric field = E; tension in the string = T; charge on the bob = q. The forces acting on the bob are shown in the figure.
At equilibrium,
T sin θ = Eq; Tcosθ = mg
∴ tan θ = \(\frac{E q}{m g}\) = \(\frac{2 \times 10^4 \times 2 \times 10^{-8}}{80 \times 10^{-6} \times 9.8}\) = \(\frac{25}{49}\)
or θ = tan-1\(\left(\frac{25}{49}\right)\) ≈ 27°
Now, cos θ = cos 27° = 0.89
∴ T = \(\frac{m g}{\cos \theta}\) = \(\frac{80 \times 10^{-6} \times 9.8}{0.89}\) = 8.8 × 10-4 N
Example 5.
A circular copper ring of radius r, placed In vacuum, has a charge q on it. Find out the electric fields
(i) at the centre of the ring, and
(ii) on the axis of the ring at a distance x from its centre.
(iii) For what value of x would the electric field be maximum?
Solution:
The charge q would be uniformly distributed along the circumference of the circular ring. So, the linear density of charge along the ring,
λ = \(\frac{q}{2 \pi r}\)
i) The electric field at the centre of the ring due to the charge on a small element of the ring would exactly be cancelled due to the charge on the diametrically opposite small element.
The whole ring is effectively an assembly of a large number of such diametrically opposite pairs of elements. As a result, the resultant electric field at the centre would he zero.
ii) Let us take a small element of length dl on the ring at the position A [Fig.]. Charge on the element dl = λdl = \(\frac{q}{2 \pi r}\)dl. The electric field at the axial point P due to the charge on the element dl,
Its axial component is
dE cosθ = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{\lambda d l}{(A P)^2} \frac{x}{(A P)}\) = \(\frac{1}{4 \pi \epsilon_0} \cdot \frac{\lambda x d l}{(A P)^3}\)
The component of dE perpendicular to the axis = dEsinθ, For charges distributed throughout the entire ring, ΣdEsinθ = 0, due to symmetry. Therefore, the resultant axial electric field at the point P,
iii) For E to be maximum, \(\frac{d E}{d x}\) = 0
For maximum E,
\(\frac{d E}{d x}\) = 0 or, r2 – 2x2 = 0 or, x = ± \(\frac{r}{\sqrt{2}}\)
The ‘±‘ sign implies that the electric field on the axis will be maximum at a distance on either side of the ring.
Example 6.
Starting from rest, an electron of mass me and a proton of mass travel through a certain distance in a uniform electric field in times t1 and t2, respectively. Find out the ratio t2/t1, neglecting the influence of gravity.
Solution:
In a uniform electric field of intensity E, the force acting on a charge q is F = qE and its acceleration, a = \(\frac{q E}{m}\)
The distance travelled from rest in time t,
Example 7.
Calculate the radius of a charged water drop which remains just suspended in equilibrium in the earth’s electric field. Charge In the water drop is equal to that of an electron. Magnitude of the earth’s electric field is 10-2 statv cm’. [e = 4.805 × 10-10 esu of charge; g = 980 cm ᐧ s-2]
Solution:
Let the radius of the water drop be r cm and mass m g.
Here intensity of electric field,
E = 10-2 statV ᐧ cm-1 ᐧ
In equilibrium, electrical force on the charged water drop = weight of the water drop.
Example 8.
A simple pendulum consists of a small sphere of mass m suspended by a thread of length l. The sphere carries a charge q. The pendulum is placed in a uniform electric field of strength E directed vertically upwards. With what period will the pendulum oscillate if the electrostatic force acting on the sphere is less than the gravitational force?
Solution:
Let the pendulum be at an angle of θ with the vertical.
Net force acting on the bob = mg – qE
Torque experienced by the bob about the point of suspension,
\(\tau\) = -(mg – qE)lsinθ
(where, the negative sign indicates that the torque experienced is restoring torque)
As the amplitude of oscillation is small,
sinθ ≈ θ
Then, \(\tau\) = -(mg – qE)θ
Again, \(\tau\) = moment of inertia (I) × angular acceleration (α)
Example 9.
Find the electric field Intensity at the centre of a semi-circular arc of radius r, uniformly charged with a charge q.
Solution:
In the Fig., the length of the semicircular wire is π r.
Linear charge density, λ = \(\frac{q}{\pi r}\)
Let the charge of a small part dl of the wire be dq.
∴ dq = λ ᐧ dl = \(\frac{q}{\pi r}\) ᐧ dl
Field intensity at the centre O due to small part,
Now the field intensity dE is resolved into two components, one along the radius which is dEx = dE cos θ and another perpen-dicular to the radius which is dEy = dE sin θ. Considering the whole wire, it is seen that all the dEx components get cancelled. Only the dEy components get remain.
∴ Total field intensity at O,
Example 10.
An electron of charge 1.6 × 10-19 C and mass 9.1 × 10-31 kg, travelling along the X -axis with a uniform velocity of 106 m s, enters in a uniform electric field of 103 V m-1 acting perpendicular to the X-axis. If the electric field extends over a length of 2 cm along the X-axis, what will be the deflection of the
electron along the direction of the field when it emerges from it?
Solution:
Here, electric field E = 103 V/m is directed along Y-axis and effective up to a length L = 2 cm along X-axis [Fig.]. An electron is projected from O with an initial speed
of vx = 106 m/s along + X-axis. Charge of the electron, e = 1.6 × 10-19 C and mass of the electron, m = 9.1 × 10-31 kg.
If we consider the motion of the electron along Y -axis, the force on the electron due to the electric field, Fy = eE.
∴ Acceleration of the electron along Y-axis, ay = \(\frac{e E}{m}\)
Let the deflection of the electron when it is emitted from the electric field be y and the time taken to cover that vertical dis-tance be t.
∴ y = \(\frac{1}{2} \cdot a_y t^2\)
or, y = \(\frac{1}{2} \cdot \frac{e E}{m} t^2\) ….. (1)
Along X -axis, force on the electron, Fx = 0
∴ Acceleration, ax = 0
∴ Velocity, vx = 106 m/s (constant)
Now distance covered along X -axis in time t is L = vxt
∴ t = \(\frac{L}{v_x}\) …… (2)
From equation (1) and (2),
y = \(\frac{1}{2} \cdot \frac{e E}{m} \cdot \frac{L^2}{v_x^2}\)
= \(\frac{1}{2} \times \frac{1.6 \times 10^{-19} \times 10^3 \times(.02)^2}{9.1 \times 10^{-31} \times\left(10^6\right)^2}\) = 0.0351 m
= 3.51 cm
Thus, required deflection is 3.51 cm