
Calculate the potential difference between 2 terminals of the battery, if $100$ Joules of work is required to transfer $20\,C$ of charge from one terminal of the battery to the other.
Answer
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Hint: We know that potential difference is the work done per unit charge and this relation is given by \[W = qV\]. In this question, this is all that is needed, we substitute the values of work done and charge to get the potential difference.
Complete step by step answer:
Let us look at the definition of potential difference : Potential difference is defined as work done to move unit positive charge from one point to another point. We can use this definition to solve the above question.
Given: Work done to transfer the charge is 100 J and the charge given is 20 C.
Hence substituting these in the equation
\[W = qV\]
\[\Rightarrow 100 = 20V\]
\[\Rightarrow \dfrac{{100}}{{20}} = V\]
\[\therefore V = 5V\].
Hence the potential difference between the terminals of the battery is 5 V.
Additional information: In general if a unit positive charge is given a displacement \[\delta l\] in a direction at an angle \[\theta \]. Workdone is given by \[\delta w = - E\delta l\cos \theta \] where $E$= electric field.
Note: Work Done in moving a charged particle from one point to another in an electrostatic field depends only on initial and final points, and not the path covered in between the 2 point;\[W = {V_f} - {V_i}\] Where \[{V_f}\] is final potential and \[{V_i}\] is initial potential and \[W\] is work done. Remember that potential due to positive charge is positive and potential due to negative charge is negative.
Complete step by step answer:
Let us look at the definition of potential difference : Potential difference is defined as work done to move unit positive charge from one point to another point. We can use this definition to solve the above question.
Given: Work done to transfer the charge is 100 J and the charge given is 20 C.
Hence substituting these in the equation
\[W = qV\]
\[\Rightarrow 100 = 20V\]
\[\Rightarrow \dfrac{{100}}{{20}} = V\]
\[\therefore V = 5V\].
Hence the potential difference between the terminals of the battery is 5 V.
Additional information: In general if a unit positive charge is given a displacement \[\delta l\] in a direction at an angle \[\theta \]. Workdone is given by \[\delta w = - E\delta l\cos \theta \] where $E$= electric field.
Note: Work Done in moving a charged particle from one point to another in an electrostatic field depends only on initial and final points, and not the path covered in between the 2 point;\[W = {V_f} - {V_i}\] Where \[{V_f}\] is final potential and \[{V_i}\] is initial potential and \[W\] is work done. Remember that potential due to positive charge is positive and potential due to negative charge is negative.
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