Chapter 17, Problem 15P

To determine

The value of $Q$ .

Answer to Problem 15P

The value of $Q$ is $2.1\times {10}^{-9}\text{C}$ .

Explanation of Solution

Given info:

The distance is $r=15.0\text{cm}$ .

The potential is $V=+125\text{V}$ .

Formula used:

The electrical potential at a point due to a charge $q$ is,

  $V=\frac{kQ}{r}$

The Coulomb’s constant is $k$ and its value is $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^2$ .

Calculation:

Substituting the given values

  $\begin{array}{c}+125\text{V}=\frac{\left(9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(Q\right)}{\left(15\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ Q=2.1\times {10}^{-9}\text{C}\end{array}$

Conclusion:

Thus, the value of $Q$ is $2.1\times {10}^{-9}\text{C}$ .