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Physics 133 Electromagnetism
Electric Potential MARLON FLORES SACEDON
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β β electric potential - What is Electric Potential?
An electric potential, the amount of work needed to move a unit charge from a reference point to a specific point against an electric field. From definition of electric field, πΈ= ππ π 2 , substitute in Eq.3β¦ becomes, From: π=πΉππππ π π=πΉπ π= ππ π 2 π π=πΉπ πΈ π=ππΈπ Defβn of Electric potential energy Eq.1 π= ππ π Electric potential for a single source charge π= π π β π Eq.2 π Eq.2 is the definition of electric potential. Where: π is electric potential energy, π is unit charge, & π is electric potential. π=π π 1 π π 2 π 2 +β¦ π π π π Electric potential for multiple source charge πΉ Substitute Eq.1 in Eq.2 β π π= ππΈπ π +π - π π=πΈπ Eq.3
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β β electric potential - Units:
If the charge π equals the magnitude π of the electron charge, 1.602π₯ 10 β19 πΆ, and the potential difference is π ππ =1 π, the change in energy is called Electron Volt. MKS: 1 π½ππ’ππ πππ’π =1 π½ πΆ =1 π£πππ‘=1 π CGS: 1 πππ π π‘ππ‘πΆ =1 π π‘ππ‘π πΈ 1 π=300 π π‘ππ‘π 1 eV=1.60π₯ 10 β19 π½ πΉ +e=1.602π₯ 10 β19 πΆ β π +π π β π - π
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electric potential Problem 1: An electric dipole consists of point charges π 1 =+12ππΆ and π 2 =β12ππΆ placed 10.0 cm apart. Compute the electric potentials at points π, π, and π. From the defβn of electric potential π π =9π₯ π. π 2 πΆ π₯ 10 β9 πΆ 4 π₯10 β2 π + β12 π₯10 β9 πΆ 14 π₯10 β2 π π=π π 1 π π 2 π 2 +β¦ π π π π π π =1, π π π =π π 1 π π 2 π 2 π π =9π₯ π. π 2 πΆ π₯ 10 β9 πΆ 13 π₯10 β2 π + β12 π₯10 β9 πΆ 13 π₯10 β2 π π π =9π₯ π. π 2 πΆ ππΆ 6ππ + β12ππΆ 4ππ π π =0 π π π =9π₯ π. π 2 πΆ π₯ 10 β9 πΆ 6 π₯10 β2 π + β12 π₯10 β9 πΆ 4 π₯10 β2 π =9π₯ π. π 2 πΆ 2 β1π₯ 10 β7 πΆ π π π =β900 π.π πΆ =β900 π½ πΆ =β900 π£πππ‘π =β900 π
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potential Difference β β What is potential difference?
The potential difference between two points is defined as: Potential difference between two points in a circuit is the work done in moving unit charge from one point to the other. The units for potential difference are Joules per coulomb, or volts. (1 π£πππ‘ = 1 π½ππ’ππ/πππ’ππππ). Potential difference is often time called Voltage or Electromotive force π=ββπ work done by electric potential π ππ =β π π β π π π ππ =β π π + π π High potential Low potential β a π ππ = π π β π π π π ππ = ππ π β ππ π How much work done by the moving charge? π ππ = π π π β π π β b π ππ π = π π β π π πΉπ π = π π β π π ππΈπ π = π π β π π πΈπ= π π β π π Note: π π β π π is potential difference or voltage
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potential Difference β β What is potential difference?
The potential difference between two points is defined as: Potential difference between two points in a circuit is the work done in moving unit charge from one point to the other. The units for potential difference are Joules per coulomb, or volts. (1 π£πππ‘ = 1 π½ππ’ππ/πππ’ππππ). Potential difference is often time called Voltage or Electromotive force π=ββπ work done by electric potential π ππ =β π π β π π π ππ =β π π + π π High potential Low potential β a π ππ = π π β π π π π ππ = ππ π β ππ π How much work done by the moving charge? π ππ = π π π β π π β b π ππ π = π π β π π πΉπ π = π π β π π ππΈπ π = π π β π π πΈπ= π π β π π Note: π π β π π is potential difference or voltage
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potential Difference Problem: In figure, a dust particle with mass π=5.0π₯ 10 β9 ππ=5.0 ππ and charge π=2.0 ππΆ starts from rest and moves in a straight line from point π to point π. What is its speed π£ at point π. The force acts on dust particle is a conservative force. So, from conservation of energyβ¦ πΈ π = πΈ π π π =9π₯ π. π 2 ππ π₯ 10 β9 πΆ 0.01π + β3π₯ 10 β9 πΆ 0.02π =1350 π πΎ π + π π = πΎ π + π π 0+ ππ π = 1 2 π π£ 2 + ππ π π π =9π₯ π. π 2 ππ π₯ 10 β9 πΆ 0.02π + β3π₯ 10 β9 πΆ 0.01π =β1350 π 1 2 π π£ 2 = ππ π β ππ π π π β π π =1350β β1350 =2700 π π£= 2π π π β π π π π£= 2 2π₯ 10 β9 πΆ π 5π₯ 10 β9 ππ =46 π π
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General problems 1. A charge of 20π₯ 10 β8 πΆ is 20 ππ from another charge of 180π₯ 10 β8 πΆ a) Find the force between the two b) What is the potential at the point which is exactly midway between the two c) What is the electric field intensity at the same point.
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