An isolated conducting sphere has a 10 cmradius. One wire carries a current of 1.000 0020Ainto it. Another wire carries a current ofout of it. How long would it take for the sphere to increase in potential by 1000V?

1. Radius of the sphere, r = 10cm or 0.10 m
2. Increase in potential,$∆V=1000V$
3. Current going in,role="math" localid="1661410034948" $i_{in}=1.0000020A$
4. Current going out,$i_{out}=1.0000000A$

Electric potential is the amount of energy required to move a unit charge from one location to another in the electric field. Electric current flowing through the cross-section is equal to the amount of charge flowing per unit time across the cross-section.

We have to use the formula of potential to find the expression for an increase in charge. By using the expression for an increase in charge in the formula of current, we can find the time required to increase the potential of the sphere.

Formulae:

The electric potential at a point due to a charge, $V=\frac{q}{4\pi {\epsilon }_{0}r}$ ...(i)

The current flowing through the cross-sectional area, $i=\frac{q}{t}$ ...(ii)

Suppose,charge on the sphere is increased by an amount $∆q$in time $∆t$, then its increase incharge due to the potential increase by an amount$∆V$that can be given by using equation (i) as follows:

$∆q=4{\epsilon }_{0}r∆V$ …(iii)

r is the radius of the sphere.

We have current going in$\left(i_{in}\right)$and going out$\left(i_{out}\right)$of the sphere, then the current change is given by,

$∆i=i_{in}-i_{out}$

Negative sign shows the current going in the opposite direction.

Thus, the time difference can be given using equation (ii) and the above value can be given as:

$∆t=\frac{∆q}{i_{in}-i_{out}}$ …(iv)

Substituting the value of$∆q$from equation(iii)in equation(iv), we get,

$$\begin{gathered} ∆\mathrm{t}=\frac{4{\mathrm{\epsilon }}_0\mathrm{r}∆\mathrm{V}}{{\mathrm{i}}_{\mathrm{in}}-{\mathrm{i}}_{\mathrm{out}}} \\ =\frac{\left(0.10\mathrm{m}\right)\left(1000\mathrm{V}\right)}{\left(8.99\times {10}^9\mathrm{F}/\mathrm{m}\right)\left(1.0000020\mathrm{A}-10000000\mathrm{A}\right)} \\ =\frac{100}{17980} \\ =5.56\times {10}^{-3}\sec \\ \approx 5.6\times {10}^{-3}\sec \end{gathered}$$

Hence, the value of the time is$5.6\times {10}^{-3}\sec$ .