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The relation between electric field and electric potential can be expressed as – “Electric field is the negative space derivative of the electric potential.” Electric field and electric potential are important concepts in the field of electrostatics in physics. The relation between electric field and electric potential is that the potential is a property of the field expressing the action of the field upon an object or a body. An electric field exists if and only if there is a difference in electric potential. If the electric charge is constant at all the points.
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Key Terms: Electric Field, Electric Potential, Potential Difference, Voltage, Charge, Equipotential Surface, Capacitance, Electric Field Intensity
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Relation Between Electric Field And Electric Potential
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The relation between electric field and electric potential is the same as that between the gravitational field and gravitational potential. The Electric Field is the negative gradient of the Electric Potential. Mathematically, the relation between electric field and electric potential or relation between e and v can be expressed as –
| \(E = - \frac{dV}{dx}\) |
where
- E is the Electric field
- V is the Electric potential
- dx is the path length.
- – Sign is the electric gradient
| Frequently Asked Questions Ques. Explain the significance of the relation between electric field and potential. (1 Mark) Ans. Relation between electric field and potential is given by – where the electric field is the negative gradient of the electric potential. Ques. What is the relationship between electric field and electric charge? (1 Mark) Ans. Electric field is an alteration of space created by presence of an electric charge. Electric field mediates electric force between a source charge and a test charge. Electric field is a vector that points away from positive charges and toward negative charges. |
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What is Electric Field?
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Electric field or electric field intensity can be stated as a force experienced by a unit positive test charge and is symbolised as E. The electric field is like any other vector field that is, it exerts a force based on a stimulus, and has units of force times inverse stimulus. In the electric field, the stimulus is given as a charge, and thus the units are NC-1. In other terms, the Electric field is a measure of force per unit charge.
Electric Field Lines
Direction of Electric Field
- The field when directed from lower potential to higher than the direction is taken as positive.
- The field when directed from higher potential to lower than the direction is taken as negative.
What is Electric Potential?
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The Electric potential is defined as the work done to move a unit charge opposite to that of an electric field or the electric potential difference is the work done by conservative forces to move a unit positive charge and is symbolised by V.
Also, the electric potential at a point is the quotient of the potential energy of any charged particle at that location divided by the charge of that particle. Its unit is JC-1. Hence, the electric potential is an estimate of energy per unit charge.
In terms of units, the electric potential and charge are related closely. They share a common factor of C-1, whereas force and energy differ only by a factor of distance that is energy is the product of force times distance.
Electric Potential
Read More: To compare the emf of two given primary cells using a potentiometer experiment
Potential Difference
The potential difference or voltage can be stated as the difference in electric potential energy between two points. In simple terms, the amount of work done in moving a unit charge from one point to another point is known as the potential difference. The potential difference is denoted as V and has the unit of volts or joules/Coulomb.
Electric Potential Difference: Electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy.
Now, Potential difference is independent of the path taken from one point to the other. It is measured by instruments such as the voltmeter, the potentiometer, and the oscilloscope. It is commonly measured in circuits, and in such situations can be calculated by using ohm’s law.
Potential Difference In A Static Field
From the above figure, it is clear, that when a charge q moves from A to B, the potential difference is unconstrained by the path taken.
Read More: NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance
Electric Field And Electric Potential Relation
The relation between e and v is given as –
| Test Charge | Formula | Electric Gradient |
|---|---|---|
| Positive | \(w \over q_0 = \int\limits_a^b \vec{E}.\vec{dl} = V_b - V_a\) | Will be higher as you go closer to test charge |
| Negative | \(w \over q_0 = \int\limits_a^b \vec{E}.\vec{dl} = V_a - V_b\) | Will be higher as you go move away from the test charge |
| Equipotential Surface | \(w \over q_0 = \int\limits_a^b \vec{E}.\vec{dl} = 0\) | Here, Electric potential is perpendicular to Electric field lines |
Derive the Relation between Electric Field And Electric Potential
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To derive a relation between electric field and electric potential, Work done in moving the test charge q0 from point a to b is given by. Therefore the relation between e and v:
\(W \displaystyle (q_{0}) _{a\rightarrow b} = \int\limits_a^b \vec{F}.\vec{dl} = q_o \int\limits_a^b \vec{E}.\vec{dl} \)
Where
- F is the applied force
- dl is the short element of the path while moving it from point a to b.
The force can be expressed as charge times electric field
\(= q_o \int\limits_a^b \vec{E}.\vec{dl} \)
Now divide both sides by test charge q0
\( {w \over q_0} = \int\limits_a^b \vec{E}.\vec{dl} \)
Now work done by the test charge is the potential Va-Vb
\( \int\limits_a^b \vec{E}.\vec{dl} = V_a - V_b\)
And Va = Vb for equipotential surface, hence relation between e and v –
| \( \int\limits_a^b \vec{E}.\vec{dl} = 0\) |
So, this is the simple derivation of the relation between electric field and electric potential.
Let’s Learn Important Cases in this Derivation
Case 1: If test charge is positive, then from the relationship between the electric field and electric potential, the potential gradient will be closer to the charge.
Case 2: If the test charge is negative, potential gradient will increase as it moves away from the test charge.
Case 3: For an equipotential surface, the potential at every point on the surface will be the same. Therefore, the potential gradient will be zero. Electric potential becomes perpendicular to the electric field lines.
| Solved Example on Relation Between Electric Field and Electric Potential Derivation Example: The Electric Potential V at any point X, Y, and Z in space is given by V=3x2 Volts. Fine the Electric Field at Any Point (2,1,2) is. Solution: Given, potential at any point X, Y, Z. ⇒ V=3x2 We have to find the electric field E at (2,1,2). We know that, ⇒ \(E = - \frac{dV}{dx}\) ⇒ \(E = \frac{-d(3x^2)}{dx} = 6x\) Thus, the electric field E at (2,1,2) is 12V/m. |
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Things To Remember
- The electric field is an estimate of force per unit charge; the electric potential is assessed as energy per unit charge.
- Potential is a property of the field that describes the action of the field upon a body.
- Potential difference or voltage is the difference in electric potential energy between two points and denoted as V with the unit as volt or J/C.
- For one point charge, the potential will remain uniform for all points a certain radial distance away. Multiple points of the same potential are called equipotential.
- For points outside a conductor, the potential is non-zero and can be calculated according to the field and the distance from the conductor.
- Positive test charges will move in the direction of the field; negative charges will move direction opposite to the field.
- Potential due to charge ‘q’ at its own location is not defined, it is infinite.
Previous Year Questions
- A sphere of radius R has a volume density of charge… [WBJEE 2013]
- Which is not the unit of electric field… [JCECE 2006]
- A cube of side 'a' has point charges +Q located at each of its vertices except… [JEE Main 2021]
- Top of the stratosphere has an electric field E (in units of Vm−1) nearly… [DUET 2009]
- A hollow insulated conduction sphere is given a positive charge… [VITEEE 2017]
- An infinitely long thin straight wire has uniform charge density of… [KCET 2020]
- The inward and outward electric flux from a closed surface are respectively… [KCET 2003]
- The magnitude of point charge due to which the electric field 30 cm… [KCET 2018]
- Two identical charges repel each other with a force equal to… [KCET 2007]
- An electron, placed in an electric field, experiences a force F of… [KEAM]
- Four point charges (with equal magnitude of charge of 5C; but with different signs… [KEAM]
- Find the correct diagram of electric lines of forces for negative charge…
- If an insulated non-conducting sphere of radius R has charge density…
- Two point charges q1(√10μC) and q2(−25μC) are placed on the x-axis… [JEE Main 2019]
Sample Questions
Ques. State how the gradient varies with the movement of charge. (3 Marks)
Ans. The gradient varies as per the situation stated below:
- Case 1: If the test charge is positive, then form the relationship between the electric field and electric potential, the potential gradient will be more near the charge.
- Case 2: If the test charge is negative, the potential gradient will move away from the test charge.
- Case 3: In the case of an equipotential surface, the potential at every point on the surface will be constant thus the potential gradient will be zero. The electric potential will be perpendicular to the electric fields.
Ques. Two charges 6 x 10-8 C and -3 x 10-8 C are located 14 cm apart. At what point or points on the line joining the two charges is the electric potential zero? Take the potential at infinity as zero. (3 Marks)
Ans. Let x be the point of zero potential be x at a distance r from the charge 1.
d= 14 cm
Electric potential, v = \(\frac{q_1}{4 \pi \in_0r}\)+\(\frac{q_2}{4 \pi \in_0(d-r)}\)
For v= 0,
\(\frac{q_1}{r}\) = - \(\frac{q_2}{(d-r)}\)
6 x \(\frac{10^{-8}}{ r}\)= -3 x \(\frac{10^{-8}}{ (0.14-r)}\)
r= 28 cm
if point p is outside,
v = \(\frac{q_1}{4 \pi \in_0r}\)+\(\frac{q_2}{4 \pi \in_0(r-d)}\)
here, r= 93.33 cm.
Ques. A regular hexagon of a side 10 cm has a charge of 5µC at each of its vertices. Evaluate the potential at the centre of the hexagon. (3 Marks)
Ans. Let O be the centre of the hexagon. Now, when point O is joined to the ends of the sides of the hexagon forms an equilateral triangle.
Now side, a = 10 cm, or 0.1 m.
Since at each corner of the hexagon a charge of 5 x 10-6 C is placed then,
Total electric potential at point O due to the charges at the six corners is given as,
V = 6 x \(\frac{q}{4 \pi \in_0r}\)
= 6 x 9 x 109x 5 x 10-6/0.1
= 2.7 x 106 V.
Ques. A sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2. Prove that if q1 is positive, it will flow from the sphere to the shell no matter what the charge q2 holds. (3 Marks)
Ans. Potential of the inner sphere due to charge q1 =V1= \(\frac{q_1}{4 \pi \in_0r_1}\)
Potential of the inner sphere due to the enclosed sphere, V2= \(\frac{q_2}{4 \pi \in_0r_2}\)
Thus, the total Potential of inner sphere V = V1+V2
= \(\frac{q_1}{4 \pi \in_0r_1}\)+ \(\frac{q_2}{4 \pi \in_0r_2}\)
The potential of shell V’ = \(\frac{q_2}{4 \pi \in_0r_2}\)
Potential difference = V-V’
= \(\frac{q_1}{4 \pi \in_0r_1}\)+ \(\frac{q_2}{4 \pi \in_0r_2}\)-\(\frac{q_2}{4 \pi \in_0r_2}\)
=\(\frac{q_1}{4 \pi \in_0r_1}\)
Hence, it is understood from the above that q1 is positive. We see that potential difference doesn’t depend upon q2. Hence, proved.
Ques. Two capacitors 6µF / 200 V and 1µF/60 V are connected in series. The maximum emf which can be applied is, Solve. (3 Marks)
a) 260 V
b) 30 V
c) none
d) 70 V
Ans. Option (d)
As the capacitors are connected in series hence charge will be the same.
The charge on small capacitance (1µF), Q1= 60µC
thus, we cannot apply charge more than 60µC
the potential on big capacitance is (6µF) for the same charge Q1 is Vb= 60/6=10 V
thus, the maximum emf= 60+10= 70V.
Ques. Explain what would happen if the capacitor given is 3 mm thick mica sheet of the dielectric constant =6 was inserted between the plates,
a) While the voltage supply remained connected
b) After the supply was disconnected (3 Marks)
Ans. (a) The dielectric constant of the mica sheet, k = 6-----given.
initial capacitance, C= 1.77 x 10-11 F
new capacitance, C’= k C= 6 x 1.77 x 10-11 F=106 pF
supply voltage, V = 100 V
New charge, q’ = C’ V = 6 x 1.77 x 10-9= 1.06 x 10-8 C
Potential across the plates remains 100 V.
(b) dielectric constant, k= 6 ---- given.
initial capacitance, C = 1.77 x 10-11 F
new capacitance, C’= k C= 6 x 1.77 x 10-11 F=106 pF
if the supply of the voltage is removed, then there will be no effect on the amount of charge on the plates.
Charge= 1.77 x 10-9 C
Potential across the plate is given as,
V’=q’/C’
V’ = 1.77 x 10-9/ 106 x 10-12= 16.7 V.
Ques. A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The exterior of the cylinder is earthed, and the inner cylinder is given a charge of 3.5 µC. Evaluate the capacitance of the system and the potential of the inner cylinder. (3 Marks)
Ans. Given,
Length of the co-axial cylinder = 15cm =0.15 m
Radius of the outer cylinder, R= 1.5 cm= 0.015 m
Radius of the inner cylinder, r= 1.4 cm= 0.014 m
Charge of the cylinder, q= 3.5 µC
Capacitance of the co-axial cylinder, C = 2π0 lr1r2
The capacitance of the system = 1.2 x 10-10 F.
Potential of the inner cylinder is, V= q/C
The potential of the inner cylinder = 2.92 x 104 V.
Ques. The area of the plate (+) is 125 cm2 and the area of the plate (-) is 100 cm2. They are parallel to each other and are separated by a distance of 0.5 cm. The capacity of a condenser with air as the dielectric is? (3 Marks)
Ans. When the area of the plate is different, we consider the area which is smaller to calculate the capacitor,
Hence, c = ∈0 A/ d ------ ∈0= 8.9 x 10-12 C2 N-1 M-2.
c= 8.9 x 10-12x 100 x 10-4/ 0.5 x 10-2
The capacity of a condenser with air as dielectric is, c= 17.8 pF.
Ques. Answer the following:
a) Calculate the potential at a point P due to a charge of 4 x 10-7 C distanced 9 cm away.
b) Also, obtain the work done in bringing a charge of 2 x10-9 C from infinity to point P. Does the solution depend upon the path along which the charge is brought? (3 Marks)
Ans. (a) Potential, V = Q4π0r = 9 x 109 Nm2C-2 x 4 x 10-7 C/ 0.09 m
V = 4 x 104 V
The potential at a point P due to a charge of 4 x 10-7 C located 9 cms away is 4 x 104 V.
(b) The work done, W = q x V
W = 2 x 10-9 C x 4 x 104 V = 8 x 10-5 J.
No, work done will be path independent. Any infinitesimal path can be divided into two perpendicular displacements i.e., one along ‘r’ and another perpendicular to ‘r’. Thus, the work done corresponding to later will be zero.
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