Question

$50pJ$of energy is stored in alocalid="1648664667304" $2.0cm\times 2.0\times 2.0cm$region of uniform electric field. What is the electric field strength?

Short answer

The electric field strength is$1.19\frac{kV}{m}$

Step-by-step solution

Step: 1 Given information

We have given that the $50pJ$of energy in localid="1648663993355" $2.0cm\times 2.0cm\times 2.0cm$.

We need to find the electric field strength.

Step: 2 Simplify

The formula of energy as it relates to the strength of the electric field is

$u_E=\frac{1}{2}{\in }_0{E}^2(30.28)$

Considering the formula$u_{E_{}}=\frac{energystored}{volumeinwhichitisstored}$, meaning localid="1648664098321" $u_E=\frac{5\times {10}^{-11}J}{{\left(0.020cm\right)}^3}=6.3\times {10}^{-6\frac{j}{m}}$. The reader has not converted the initial units to base units; this solution omits changing $\left(pJ\right)$ to $\left(J\right)$and omits changing $\left(c{m}^3\right)$to $\left(m^3\right)$.

If we rearrange $(30.28)$to solve for $E$

$$\begin{gathered} u_E=\frac{1}{2}{\in }_0{E}^2 \\ \to 2u_E={\in }_o{E}^2 \\ \to \frac{2u_E}{{\in }_o}=E^2 \\ \to E=\sqrt{\frac{2u_E}{{\in }_0}} \end{gathered}$$

Using the numerical values of given variable and constant

localid="1648664073878" $$\begin{gathered} E=\sqrt{\frac{2u_E}{{\in }_0}} \\ E=\sqrt{\frac{2\left(6.3\times {10}^{-6}{\displaystyle \frac{j}{m}}\right)}{8.85\times {10}^{-12}{\displaystyle \frac{c^2}{N{m}^2}}}} \\ =1.19\times {10}^3\frac{V}{m} \\ =1.19kV/m \end{gathered}$$