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Electric Potential The electric potential is a scalar quantity which is generally denoted by φ, φ_{E} or V. It is also termed as electric field potential or the electrostatic potential. Electric potential, at any point, is defined as the negative line integral of electric field from infinity to that point along any path. Electric potential is a scalar quantity. Potential difference, between any two points, in an electric field is negative line integral of electric field between them along any curve joining them together. In simply words, potential difference between any two points, in an electric field is defined ads the work done in taking a unit positive charge from one point to the other against the electric field. W_{AB} = q [V_{A}-V_{B}] So, V = [V_{A}-V_{B}] = W/q Units:- volt (S.I), stat-volt (C.G.S) Dimension:- [V] = [ML^{2}T^{-3}A^{-1}] Relation between volt and stat-volt:- 1 volt = (1/300) stat-volt The electric potential at any point is equal to the quotient of the electric potential energy of the particle and the charge of the particle at a particular point. The potential energy is measured in joules while the charge is measured in coulombs. The potential may be calculated in either a static or a dynamic electric field and as described above its unit is joules per coulomb i.e. JC^{-1} or volts V. The potential energy for a positive charge has inverse relationship with electric field. The energy increases when it moves against an electric field and falls when it moves with the electric field. For a negative charge, the situation is quite opposite. Potential at point due to several charges:- V = (1/4π ε_{0}) [q_{1}/r_{1 }+ q_{2}/r_{2} + q_{3}/r_{3}] = V_{1}+V_{2}+ V_{3}+…. Potential due to charged spherical shell:- To obtain the electric potential of a charged spherical shell, first we have to fin d out the electric filed inside and outside a thin uniformly charged spherical shell of radius R and charge Q. The charge distribution is spherical, then E is radial, and its magnitude depends only on the distance r from the center of the sphere. Therefore the Gaussian surface is the spherical surface that has the same center as the spherical shell of charge, of radius r < R inside and r> R outside. The electric flux inside the shell will be, Φ_{E} = ∫ E(r). dS = E(r) ∫ dS = 4πr^{2} E(r) As we know, the charge inside this surface is zero, thus, ΣQ_{i} = 0. In accordance to Gauss’s law, Φ_{E}(inside) = 0 That is, E(inside) = 0 So the potential remains constant at the value it reaches at the surface. Thus, V_{inside} = kQ/R = (1/4π ε_{0}) (Q/R) The flux outside is : Φ_{E}= ∫ E(r). dS = E(r) ∫ dS = 4πr^{2} E(r) The charge inside this surface is q/εo. In accordance to Gauss’s law, Φ_{E}(outside) = 4πr^{2} E(r) = Q/ε_{o} Solving for E(r), we get, E(outside) = Q/4πε_{o} (1/r^{2}) Thus elecrtric potential will be, V_{outside}=Er =[Q/4πε_{o} (1/r^{2})]r =Q/4πε_{o}r At the surface of the charged spherical shell, r = R. Therefore, V_{surface }= Q/4πε_{o}R Refer this simulation for electric potential and work We know,W=qV. I use this animation to show electric potential and contrast it with potential energy. It shows a micro sized tractor pushing charges towards a charged block. Using a "work" meter it shows the energy it takes to move the charge. Have the student conclude the relationship between the work and the number of charges. Then relate this to "W=qV." Potential due to a uniformly charged non-conducting sphere:- (a) Outside, V_{out} = (1/4π ε_{0}) (q/r) (b) Inside, V_{in} = (1/4π ε_{0}) [q(3R^{2}-r^{2})/2R^{3}] (c) On the surface, V_{surface} = (1/4π ε_{0}) (q/R) (d) In center, V_{center} = (3/2) [(1/4π ε_{0}) (q/R)] = 3/2 [V_{surface}] Potential at any point due to an electric dipole:- V (r,θ) = qa cosθ/4πε_{0}r^{2} = p cosθ/4πε_{0}r^{2} (a) Point lying on the axial line:- V = p/4πε_{0}r^{2} (b) Point situated on equatorial lines:- V = 0 Refer this video to know more about electric potential Electric potential energy of a system (a) A system of two charges The total work done in making an assembly of charges by bringing them from infinity to their respective places is known as the potential energy of the system. Potential energy is said to be positive if the charges are brought together against the forces of repulsion between them and negative if the charges move under the effect of attractive forces. Thus, the electric potential energy for point charges q_{1} and q_{2} separated by a distance r_{12}, will be, U = q_{1}q_{2}/ 4π∈_{0}r_{12} (b) A system of three charges[q_{1}, q_{2} and q_{3}] Let three charges q_{1},q_{2} and q_{3} constituting the system. These three charges from three pairs q_{1}q_{2},q_{1}q_{3} and q_{2}q_{3} respectively. The net potential energy of the system is the sum total of potential energies required to assemble these three pairs separately. Thus the net potential energy of the system will be, U = 1/ 4π∈_{0} (q_{1}q_{2}/r_{12 }+ q_{1}q_{3}/r_{13 }+_{ }q_{2}q_{3}/r_{23}) (c) Potential energy at point due to several charges If there are ‘N’ charges constituting the configuration, the total number of possible pairs for computing potential energy of the system are, 12,13,14,...., 1N, 23, 24, …..,2N, 34, 35,....,3N and so on . A general expression for the potential energy of ‘N’ charges can be written as, Here, . The factor ½ appears in the expression due to the reason that the combinations q_{j}q_{k} and q_{k}q_{j} are to be counted as one. Conceptual Question: During a thunderstorm, a lightning strike places a huge static charge on your car. Why don’t you notice this charge as long as you remain inside the car? Answer: The accumulated charge is all on the outside of the conducting car, so it will only affect you if you step outside and offer it a conducting path to the ground. Why: Since the car body is an electrical conductor and its charges are in equilibrium, the car has a uniform voltage and there is no electric field inside the car. Outside the car, however, there is a substantial electric field. If your body provides a conducting path to the ground, that electric field will push charges through you and you will experience a shock. Similarly, if electric power lines ever fall on your car during a storm, stay inside the car to avoid a potentially lethal shock.
The electric potential is a scalar quantity which is generally denoted by φ, φ_{E} or V. It is also termed as electric field potential or the electrostatic potential.
Electric potential, at any point, is defined as the negative line integral of electric field from infinity to that point along any path.
Electric potential is a scalar quantity.
Potential difference, between any two points, in an electric field is negative line integral of electric field between them along any curve joining them together.
In simply words, potential difference between any two points, in an electric field is defined ads the work done in taking a unit positive charge from one point to the other against the electric field.
W_{AB} = q [V_{A}-V_{B}]
So, V = [V_{A}-V_{B}] = W/q
Units:- volt (S.I), stat-volt (C.G.S)
Dimension:- [V] = [ML^{2}T^{-3}A^{-1}]
Relation between volt and stat-volt:- 1 volt = (1/300) stat-volt
The electric potential at any point is equal to the quotient of the electric potential energy of the particle and the charge of the particle at a particular point. The potential energy is measured in joules while the charge is measured in coulombs. The potential may be calculated in either a static or a dynamic electric field and as described above its unit is joules per coulomb i.e. JC^{-1} or volts V.
The potential energy for a positive charge has inverse relationship with electric field. The energy increases when it moves against an electric field and falls when it moves with the electric field. For a negative charge, the situation is quite opposite.
V = (1/4π ε_{0}) [q_{1}/r_{1 }+ q_{2}/r_{2} + q_{3}/r_{3}]
= V_{1}+V_{2}+ V_{3}+….
To obtain the electric potential of a charged spherical shell, first we have to fin d out the electric filed inside and outside a thin uniformly charged spherical shell of radius R and charge Q.
The charge distribution is spherical, then E is radial, and its magnitude depends only on the distance r from the center of the sphere. Therefore the Gaussian surface is the spherical surface that has the same center as the spherical shell of charge, of radius r < R inside and r> R outside.
The electric flux inside the shell will be,
Φ_{E} = ∫ E(r). dS = E(r) ∫ dS = 4πr^{2} E(r)
As we know, the charge inside this surface is zero, thus,
ΣQ_{i} = 0.
In accordance to Gauss’s law,
Φ_{E}(inside) = 0
That is, E(inside) = 0
So the potential remains constant at the value it reaches at the surface.
Thus, V_{inside} = kQ/R = (1/4π ε_{0}) (Q/R)
The flux outside is : Φ_{E}= ∫ E(r). dS = E(r) ∫ dS = 4πr^{2} E(r)
The charge inside this surface is q/εo.
Φ_{E}(outside) = 4πr^{2} E(r) = Q/ε_{o}
Solving for E(r), we get,
E(outside) = Q/4πε_{o} (1/r^{2})
Thus elecrtric potential will be,
V_{outside}=Er
=[Q/4πε_{o} (1/r^{2})]r
=Q/4πε_{o}r
At the surface of the charged spherical shell, r = R.
Therefore, V_{surface }= Q/4πε_{o}R
We know,W=qV. I use this animation to show electric potential and contrast it with potential energy. It shows a micro sized tractor pushing charges towards a charged block. Using a "work" meter it shows the energy it takes to move the charge. Have the student conclude the relationship between the work and the number of charges. Then relate this to "W=qV."
(a) Outside, V_{out} = (1/4π ε_{0}) (q/r)
(b) Inside, V_{in} = (1/4π ε_{0}) [q(3R^{2}-r^{2})/2R^{3}]
(c) On the surface, V_{surface} = (1/4π ε_{0}) (q/R)
(d) In center, V_{center} = (3/2) [(1/4π ε_{0}) (q/R)] = 3/2 [V_{surface}]
V (r,θ) = qa cosθ/4πε_{0}r^{2} = p cosθ/4πε_{0}r^{2}
(a) Point lying on the axial line:- V = p/4πε_{0}r^{2}
(b) Point situated on equatorial lines:- V = 0
(a) A system of two charges
The total work done in making an assembly of charges by bringing them from infinity to their respective places is known as the potential energy of the system.
Potential energy is said to be positive if the charges are brought together against the forces of repulsion between them and negative if the charges move under the effect of attractive forces.
Thus, the electric potential energy for point charges q_{1} and q_{2} separated by a distance r_{12}, will be,
U = q_{1}q_{2}/ 4π∈_{0}r_{12}
(b) A system of three charges[q_{1}, q_{2} and q_{3}]
Let three charges q_{1},q_{2} and q_{3} constituting the system. These three charges from three pairs q_{1}q_{2},q_{1}q_{3} and q_{2}q_{3} respectively.
The net potential energy of the system is the sum total of potential energies required to assemble these three pairs separately.
Thus the net potential energy of the system will be,
U = 1/ 4π∈_{0} (q_{1}q_{2}/r_{12 }+ q_{1}q_{3}/r_{13 }+_{ }q_{2}q_{3}/r_{23})
(c) Potential energy at point due to several charges
If there are ‘N’ charges constituting the configuration, the total number of possible pairs for computing potential energy of the system are,
12,13,14,...., 1N, 23, 24, …..,2N, 34, 35,....,3N and so on .
A general expression for the potential energy of ‘N’ charges can be written as,
Here,
.
The factor ½ appears in the expression due to the reason that the combinations q_{j}q_{k} and q_{k}q_{j} are to be counted as one.
During a thunderstorm, a lightning strike places a huge static charge on your car. Why don’t you notice this charge as long as you remain inside the car?
The accumulated charge is all on the outside of the conducting car, so it will only affect you if you step outside and offer it a conducting path to the ground.
Since the car body is an electrical conductor and its charges are in equilibrium, the car has a uniform voltage and there is no electric field inside the car. Outside the car, however, there is a substantial electric field. If your body provides a conducting path to the ground, that electric field will push charges through you and you will experience a shock. Similarly, if electric power lines ever fall on your car during a storm, stay inside the car to avoid a potentially lethal shock.
Electric potential is a analogous to height in mechanics and temperature in heat.
It is the negative line integral of electric field between infinity and the given point.
Electric potential at any point varies inversely as the distance of the point from source charge.
Electric potential at any point due to a positive charge is positive while that due to a negative charge is negative.
Electric Potential Energy is not the same as Electrical Potential.
Electrical Potential can also be described by the terms, potential difference, voltage, potential drop, potential rise, electromotive force, and EMF. These terms may differ slightly in meaning depending on the situation.
The variable we use for potential difference is V and the unit for potential difference is also V (volts). Don’t let that confuse you when you see V = 1.5V
The electron volt is not a smaller unit of the volt, it’s a smaller unit of the Joule.
A point charge q is palced at a distance of r from the center of an uncharged conducting sphere of radius R(<r). The potential at any point on the sphere is,
(a) zero (b) (1/4π ε_{0}) (q/r)
(c) (1/4π ε_{0}) (qR/r^{2}) (d) (1/4π ε_{0}) (qr^{2}/R)
Since, potential V is same for all points of the sphere. Therefore, we can calculate its value at the center of the sphere.
Thus, V = (1/4π ε_{0}) (q/r) + V ‘
V ‘=potential at center due to induced charges = 0
Because net induced charge will be zero.
Therefore, V = (1/4π ε_{0}) (q/r)
From the above observation, we conclude that, option (b) is correct.
A particle of charge q_{1} = +6.0 µC is located on the x-axis at the point x_{1} = 5.1 cm. A second particle of charge q_{2} = -5.0 µC is placed on the x-axis at x_{2} = -3.4 cm. What is the absolute electric potential at the origin (x = 0)? How much work must we perform in order to slowly move a charge of q_{3} = -7.0 µC from infinity to the origin, whilst keeping the other two charges fixed?
The absolute electric potential at the origin due to the first charge is
V_{1} = k_{e}q_{1}/x_{1}
= (8.98810^{9}) (610^{-6})/(5.110^{-2})
= 1.0610^{6}V
Likewise, the absolute electric potential at the origin due to the second charge is
= (8.98810^{9}) (-510^{-6})/(3.410^{-2})
= -1.3210^{6}V
The net potential V at the origin is simply the algebraic sum of the potentials due to each charge taken in isolation. Thus,
V = V_{1}+V_{2} = -2.6410^{5} V
W = q_{3}V
= (-710^{-6}) (-2.6410^{5}) = 1.85 J
Question: Suppose that three point charges, q_{a}, q_{b}, and q_{c}, are arranged at the vertices of a right-angled triangle, as shown in the diagram. What is the absolute electric potential of the third charge if q_{a} = -6.0 µC q_{b} = +4.0 µC q_{c} = +2.0 µC, a = 4.0 m, and b = 3.0 m? Suppose that the third charge, which is initially at rest, is repelled to infinity by the combined electric field of the other two charges, which are held fixed. What is the final kinetic energy of the third charge?
The absolute electric potential of the third charge due to the presence of the first charge is
V_{a} = k_{e}q_{a}/c
= (8.98810^{9}) (-610^{-6})/( √ 4^{2}+3^{2})
= -1.0810^{4} V,
= (8.98810^{9}) (-410^{-6})/(3)
= 1.2010^{4} V,
The net absolute potential of the third charge V_{c} is simply the algebraic sum of the potentials due to the other two charges taken in isolation. Thus,
V_{c} = V_{a}+V_{b} = 1.2010^{3}V.
The change in electric potential energy of the third charge as it moves from its initial position to infinity is the product of the third charge, q_{c }, and the difference in electric potential (-V_{c}) between infinity and the initial position. It follows that
∆P = – q_{c}V_{c}
= – (210^{-6}) (1.210^{3})
= – 2.4010^{-3}J
This decrease in the potential energy of the charge is offset by a corresponding increase ∆K = – ∆P in its kinetic energy. Since the initial kinetic energy of the third charge is zero (because it is initially at rest), the final kinetic energy is simply
K = ∆K = – ∆P = – 2.4010^{-3}J
Question 1
When we rub a glass rod with a silk cloth then
(a) glass rod acquires negative charge while silk acquires positive charge
(b) glass rod acquires positive charge while silk acquires negative charges
(c) both glass rod and silk acquire negative charge
(d) both glass rod and silk acquire positive charge
Question 2
If free space between the plates of a capacitor is replaced by a dielectric
(a) The potential difference remains constant capacitance and energy stored increases
(b) The potential difference remains constant capacitance decreases and energy increases
(c)The potential difference decreases but both capacitance and energy increase
(d) both potential difference and capacitance decrease but energy increases
Question 3
Potential difference is the work done in moving unit positive charge form one point to another
(a) in the direction of electric intensity
(b) against electric intensity
(c) in any direction
(d) in the direction of electric flux
Question 4
Equipotential planes are
(a) parallel to one another
(b) non parallel to one another
(c) intersecting
(d) circular
Question 5
A hollow metallic sphere of 8cm diameter is charged with 4 x 10 8C. The potential on its surface will be
(a) 900 volts (b) 9000 volts
(c) 90 volts (d) zero
b
c
a
You might like to refer Energy Stored in a Capacitor.
For getting an idea of the type of questions asked, refer the Previous Year Question Papers.
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