| Force $\vec F = k \frac{Qq}{r^2} \hat a _r [=] N$ | Field $|\vec E| = k\frac{Q}{r^2} \hat a_r$ | $\vec E = \frac{\vec F}{q}$ | | --- | --- | --- | | Electrostatic Potential Energy (U) $U = k \frac{Qq}{r}$ | Electric Potential (Voltage) $V = k\frac{Q}{r}$ | $V = \frac{U}{q}$ |

How is it that $q_0$ can experience the force due to $q$ (note that they are not even touching)?

• The concept of field allows us to answer the above question
  • Electric fields always move from positive to negative
• Charge $q$ modifies the space around it
  • This modification of space (and perhaps time) is called a field, in this case, an electric field $\overrightarrow{E}$
• Electric field is a vector and for the above case: $|\overrightarrow{E_q}| = \frac{|\overrightarrow{F_{q_0,q}}|}{|q_0|} = \frac{k|q|}{r^2} = \frac{1}{4 \pi \epsilon _0} = \frac{|q|}{r^2}$
  • The units for $\overrightarrow{E}$ is $[E] = [\frac{N}{C}]$
• In calculating & depicting $\overrightarrow{E}$ for charge $q$, we assume $q_0$ is positive and negligibly small

<aside> 💫 In other words $\overrightarrow{E_q} = lim_{q_0 \to 0} \frac{\overrightarrow{F_{q_0,q}}}{q_0}$

</aside>

Graphical Representation of $\overrightarrow{E}$

Charged Particles in Presence of an Electric Field

Perfect Conductors Perfect Dielectric