Solved Examples on Electric Potential Energy
Example 1: Suppose you have a 12.0 V motorcycle battery that can move 5000 C of charge and a 12.0 V car battery that can move 60,000 C of charge. How much energy does each deliver?
Solution:
Given,
The voltage of the battery is V =12.0 V.
The charge that the motorcycle battery move is Q = 5000 C.
The 12.0 V car battery can move 60,000 C of charge.
When such a battery moves charge, it puts the charge through a potential difference of 12.0 V, and the charge is given a change in potential energy equal to Δ(PE) = qΔV.
For the motorcycle battery, q = 5000 C and ΔV = 12.0 V. The total energy delivered by the motorcycle battery is
ΔPEmotorcycle = (5000 C) × (12.0 V)
ΔPEmotorcycle = 6.00 × 104 J
Now, for the car battery,
ΔPEcar = (60,000 C) × (12.0 V)
ΔPEcar =7.20 × 105 J
Example 2: A particle of mass 40 mg carrying a charge 5 × 10-9 C is moving directly towards a fixed positive point charge of magnitude 10-8 C. When it is at a distance of 10 cm from the fixed point charge, it has a velocity of 50 cm/s. At what distance from the fixed point charge will the particle come momentarily to rest? Is the acceleration constant during motion?
Solution:
Given,
The mass of the particle m = 40 mg.
The charge of the particle Q = 5×10-9 C.
The fixed positive point charge of magnitude q =10-8 C.
Velocity of the charged particle is v = 50 cm/s = 0.5 m/s
The particle comes to rest momentarily at a distance r from the fixed charge, from conservation of energy we have,
According to the law of conservation of energy, the total energy of the system = Constant
i.e., (K.E. + P.E.) = constant.
The expression for the kinetic energy can be expressed as,
[Tex]K.E= \frac{1}{2}mu^2 [/Tex]
The expression for the potential energy can be expressed as,
Now,
(1/2)mu2 + (1/4πεo) × [Qq/a] = (1/4πεo) × [Qq/r]
(1/2)mu2 = (1/4πεo) [Qq/r – Qq/a]
(1/2)mu2 = (1/4πεo) Qq[1/r – 1/a]
Substituting the values in the above equation,
1/2 × 40 × 10-6 × (0.5)2= 9 × 109 × 10-8 × 5 × 10-9 × [ 1/r – 1/(10 × 10-2)]
or, [1/r – 10] = (5×10-5)/(9×5×10-8) = 100/9
or, 1/r = (100/9) + 10
or, 1/r = 190/9 m
or r = 4.7 × 10-2 m
Since, F = [1/4πεo] × [Qq/r2]
Therefore, acceleration = F/m ∝ 1/r2 , i.e., acceleration is not constant during motion.
Example 3: A ball of mass 5 g and charge 10-7 C moves from point A, whose potential is 500 V, to point B, whose potential is zero. What is the velocity of the ball at point A if at point B, it is 25 cm per second?
Solution:
Given,
The mass ball is 5 g.
The charge of the particle is 10-7 C.
The potential of ball at point A is 500 V and potential at point B is zero.
Suppose u be the velocity of the ball at point A.
The work done on the charge by the field given by,
W = q × (VA – VB)
Substitute the value in the above expression,
W = 10-7 C× (500 V – 0 V)
W = 5 × 10-5 J
Therefore,
W = (1/2) mv2 – (1/2) mu2
5 × 10-5 J= (1/2) × (5/1000 )×[(1/4)2 – u2]
2 × 10-2 = 1/16 – u2
u2 = (1/16) – 0.02
u2 = (1- 0.32)/16
u2 = 0.0425
Therefore, u =0.206 m/s
u = 20.6 cm/s.
Example 4: When a 12.0 V car battery runs a single 30.0 W headlight, how many electrons pass through it each second?
Solution:
The expression for the potential energy can be written as,
Δ(PE) = qΔV
Rearrange the above expression,
q = Δ(PE)/ΔV
Substitute the values in the above equation,
q = −30.0 J/ 12.0 V
q = −30.0 J/ 12.0 J/C
q = −2.50 C
The number of electrons n can be calculated as,
n = q/e
n = −2.50 C/(−1.60 × 10−19 C/e)
n = 1.56 × 1019 electrons
Example 5: How much work is required to be done in order to bring two charges of magnitude 3 C and 5 C from a separation of infinite distance to a separation of 0.5 m?
Solution:
Given,
Two charges of magnitude 3 C and 5 C.
The separation between two charges are 0.5 m.
The potential at P due to the charge Q can be expressed as
[Tex] U_{r} =\frac{Qq}{4\pi\epsilon_0r}[/Tex]
∆U = U0 – Ur
∆U = 0 J – [-(9 × 109 Nm2/C2× 5 C × 3 C)/0.5 m] J = 2.7 × 1011J.
Therefore, ∆U = 2.7 × 1011 J.
Electric Potential Energy
Electrical potential energy is the cumulative effect of the position and configuration of a charged object and its neighboring charges. The electric potential energy of a charged object governs its motion in the local electric field.
Sometimes electrical potential energy is confused with electric potential, however, the electric potential at a specific point in an electric field is the amount of work required to transport a unit charge from a reference point to that specific point and electrical potential energy is the amount of energy required to move a charge against the electric field.
In this article, let’s understand the electrical potential energy, electric potential, their key concepts, applications, and solved problems.
Table of Content
- What is Electric Potential Energy?
- Electric Potential Energy Formula
- Electric Potential Energy of a Point Charge
- Electric Potential Energy of a System of Charges
- What is Electric Potential?
- What is Electric Potential Difference?
- Electric Potential Derivation
- Electric Potential of a Point Charge
- Solved Examples on Electric Potential Energy
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