
Which of the following represent voltage?
A. $\dfrac{workdone}{current\times time}$
B. $\dfrac{Workdone\times Time}{Current}$
C. \[Workdone\times Charge\]
D. \[Workdone\times Charge\times Time\]
Answer
578.7k+ views
Hint: Charges under influence of electric force experience force. Use coulomb’s law. Convert charge into current. Both quantities are different. SI unit of current is ampere and SI unit of charge is coulomb.
Complete step by step answer:
Electric potential is named as voltage. SI Unit of electric potential is joule/coulomb.
Electric potential at a point in the electric field is defined as the amount of work done to bring a unit positive charge from infinity to that point against the direction of electric intensity.
Let q1 and q2 be the two charges separated by a distance r. Consider both the charges are the same therefore q1=q2=q.
If ${{q}_{0}}$is the reference point then work done to bring the charge q from a reference point to r point and it is given by.
$W=\dfrac{q\times q}{4\pi {{\varepsilon }_{0}}r}$ -----------------(1)
We know that, voltage is given by
Voltage = electric field *distance between two points.
$\begin{align}
& V=E*r \\
& V=\dfrac{F}{q}*r \\
\end{align}$------------------------------(2)
By Coulomb’s law, Force is given by
$F=\dfrac{q\times q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$
Now put above value F in equation (2), we get
$\begin{align}
& V=\dfrac{\dfrac{q\times q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}}{q}*r \\
& V=\dfrac{\dfrac{q\times q}{4\pi {{\varepsilon }_{0}}r}}{q} \\
& V=\dfrac{W}{q} \\
\end{align}$ Compare with work done equation (1)
Voltage = $\dfrac{workdone}{\text{charge}}$ ---------------------------(3)
We know that current is equal to charge per unit time.
Current = $\dfrac{\text{charge}(q)}{time}$
Charge (q) = $current*time$
Put above value in equation (1)
Voltage = $\dfrac{workdone}{current\times time}$
Therefore option A is correct.
Note: In this kind of question you can use many ways. You can get this formula using Coulomb's law or using only electric fields. Students may forget the conversion of charge into current. Make sure to use this relationship. Like this you we get the correct option.
Complete step by step answer:
Electric potential is named as voltage. SI Unit of electric potential is joule/coulomb.
Electric potential at a point in the electric field is defined as the amount of work done to bring a unit positive charge from infinity to that point against the direction of electric intensity.
Let q1 and q2 be the two charges separated by a distance r. Consider both the charges are the same therefore q1=q2=q.
If ${{q}_{0}}$is the reference point then work done to bring the charge q from a reference point to r point and it is given by.
$W=\dfrac{q\times q}{4\pi {{\varepsilon }_{0}}r}$ -----------------(1)
We know that, voltage is given by
Voltage = electric field *distance between two points.
$\begin{align}
& V=E*r \\
& V=\dfrac{F}{q}*r \\
\end{align}$------------------------------(2)
By Coulomb’s law, Force is given by
$F=\dfrac{q\times q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$
Now put above value F in equation (2), we get
$\begin{align}
& V=\dfrac{\dfrac{q\times q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}}{q}*r \\
& V=\dfrac{\dfrac{q\times q}{4\pi {{\varepsilon }_{0}}r}}{q} \\
& V=\dfrac{W}{q} \\
\end{align}$ Compare with work done equation (1)
Voltage = $\dfrac{workdone}{\text{charge}}$ ---------------------------(3)
We know that current is equal to charge per unit time.
Current = $\dfrac{\text{charge}(q)}{time}$
Charge (q) = $current*time$
Put above value in equation (1)
Voltage = $\dfrac{workdone}{current\times time}$
Therefore option A is correct.
Note: In this kind of question you can use many ways. You can get this formula using Coulomb's law or using only electric fields. Students may forget the conversion of charge into current. Make sure to use this relationship. Like this you we get the correct option.
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