Question

The electric field 2.0 cm from a small object points away from the object with a strength of 270,000 N/C. What is the object’s charge?

Short answer

$Q=12.0\times {10}^{-9}\mathrm{C}$

Step-by-step solution

Given information

It is given that,

Electric field strength,$E=270000N/C$

Distance from a small object,$r=2.0cm=0.02m$

Explanation

Electric field at a point is given by : $E=\frac{kQ}{r^2}$

Q is the charge on an object

localid="1650565043933" $\begin{array}{r}Q=\frac{E{r}^2}{k}\\ Q=\frac{270000\mathrm{N}/\mathrm{C}\times (0.02\mathrm{m}{)}^2}{9\times {10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2}\\ Q=12.0\times {10}^{-9}\mathrm{C}\end{array}$

So, the charge on the object is localid="1650565055607" $12.0\times {10}^{-9}\mathrm{C}$. Hence, this is the required solution.