Question

What is the change in electric potential energy if a -5.62x10 4C charge is moved 0.357min the direction of a 3880N/C field?6.11J0.778J-0.778J-6.11J

Answer 1

Electric potential is the work done in bringing unit positive charge .

Here work(w) is given by

`\begin{gathered} W=\text{ FS}\cos\theta;\begin{cases}F={force=\text{ 3880N}} \\ s={displacement=\text{ 0.357m}}\end{cases} \\ \therefore W=\text{ 3880}*0.357\begin{cases}\theta={0\text{ \lparen for same direction\rparen}} \\ \cos\theta={1}\end{cases} \\ \therefore W=1385.16\text{ J} \end{gathered}`

Again

Total change in electric potential energy (V) is given by

`\begin{gathered} V=\text{ W}* Q\begin{cases}V={electric\text{ potential}} \\ Q={charge=-5.62*10^{-4^{\placeholder{⬚}}}}\end{cases} \\ V=\text{ 1385.16}*-5.62*10^(-4)\text{ = -0.778J} \end{gathered}`

Final answer is -0.778J