Potential Difference

Concept Explanation

Potential Difference

Electric Potential:

Electric potential at a point in electric field is defined to be equal to the minimum work done by an external agent in moving a unit positive charge from infinity or a reference point to that point against the electrical force of the field. If W is the work done by external agent in bringing a positive test charge q_o from infinity to a point then the potential V at that point,

$large V=frac{W_{ext}}{q_{0}}$

According to Newtons Third Law of motion i.e. for every action there is equal and opposite reaction

The work done by an external agent in bringing a positive test charge q_o from infinity to a point will be equal and opposite ti the work done by the electric field on a positive test charge q_{o, if}

$W_{E}=work;done;by;electric field$

$W_{E}= -W_{Ext}$

$V=-frac{W_{E}}{q_{0}}$

The SI unit of electric potential is joule/coulomb or volt.

Potential Difference: Potential difference between two points is equal to the minimum work done in moving a unit positive test charge from one point to the other.

$V_{B}-V_{A}=frac{W_{AB}}{q_{0}}Rightarrow W_{AB}=q_{0}(V_{B}-V_{A})$ ,

$W_{E}$ is the work done by the electric field then

$W_{AB.}=-W_{E}$

$W_{E.}=-W_{AB}= -q_0(V_B-V_A)$

$W_{E.}=q_0(V_A-V_B)$

Hence the work done by external agent

$W_{AB}=q_{0}(V_{B}-V_{A})$

And Work done by the electric field is

$W_{E.}=q_0(V_A-V_B)$

We know that heat energy flows from the higher temperature to the lower temperature, and this process stops when the both are at the same temperature. Same is the case with electricity, when the ends of an electrical conductor are at different potentials or we can say that when there is a potential difference - charge in the conductor flow from the higher potential to the lower potential.The flow of charge persists until both ends reach the same potential. Without a potential difference, no flow of charge will take place.

Illustration: How much work is done in moving a charge of 2 C from a point of 118 V to a point at 128 V ?

Solution: Given, charge, q = 2 C

Potential at point A

$V_{A}=118;V$

and Potential at point B

$V_{B}=128;V$

Work done, W = ? We know that,

Potential difference,

$Delta V=V_{B}-V_{A}$ $=128-118=10V$

$large therefore$ Work Done,

$W_{AB}=q_{0}(V_{B}-V_{A})$

$W_{AB}=q_{0};Delta V$

$W=Delta Vtimes q=10times2=20 J$

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Sample Questions

(More Questions for each concept available in Login)

Question : 1

A suitable unit for expressing electric field strength is __________________
A. V/C	B. C/ $m^2$	C. Am	D. N/C
Right Option : D
View Explanation

Question : 2

A graph is plotted between the potential difference (applied across the ends of a conductor) and the current (flowing through the conductor ). The graph is a straight line __________________.
A. Having intercepts on both axes	B. Having in intercept on the X - axis	C. Having an itercept on the Y - axis	D. None of these
Right Option : D
View Explanation

Question : 3

Unit of electromotive force is ___________ .
A. Volt	B. Second	C. Metre	D. $m ; s^-^1$
Right Option : A
View Explanation

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Potential Difference

Potential Difference

Electric Potential:

Students / Parents Reviews [10]

Aman Kumar Shrivastava

Ayan Ghosh

Cheshta

Manish Kumar

Shivam Rana

Archit Segal

Prisha Gupta

Hiya Gupta

Eshan Arora

Shreya Shrivastava