Question

How much work is required to carry a $6\mu \text{C}$ charge from the negative terminal of a $9\text{V}$ cell to the positive terminal?

• A. $54\times {10}^{-3}\text{J}$
• B. $54\times {10}^{-9}\text{J}$
• C. $54\times {10}^{-6}\text{J}$
• D. $54\times {10}^{-12}\text{J}$

Answer 1

The correct option is C $54\times {10}^{-6}\text{J}$

E.M.F of the cell is $9\text{V}$.

From the definition of E.M.F of cell,

$E=\frac{W}{q}$

Where $W$ is work done by the cell in taking the $+ve$ charge from one terminal to the other.

$\Rightarrow 9\text{V}=\frac{W}{6\mu \text{C}}$

$\Rightarrow W=9\times 6=54\mu \text{J}$

$\therefore W=54\times {10}^{-6}\text{J}$

Hence, option $\left(c\right)$ is the correct.

$$\begin{array}{c}\text{Why this question?}\\ \text{Emf of the cell is the work done}\\ \text{by cell in moving a unit positive}\\ \text{charge in closed circuit i.e, from its one}\\ \text{terminal to the another one.}\end{array}$$