A charge of $- 2 C$ $-2 C$ is at the origin. How much energy would be applied to or released from a $4 C$ $4 C$ charge if it is moved from $(- 7, 5)$ $(-7 ,5 )$ to $(9, - 2)$ $(9 ,-2 )$ ?

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Answer 1 · 2018-01-25T17:40:12

The energy released is $= 0.56 \cdot 10^{9} J$ $=0.56*10^9J$

The potential energy is

$U = k \frac{q_{1} q_{2}}{r}$ $U=k(q_1q_2)/r$

The charge $q_{1} = - 2 C$ $q_1=-2C$

The charge $q_{2} = 4 C$ $q_2=4C$

The Coulomb's constant is $k = 9 \cdot 10^{9} N m^{2} C^{- 2}$ $k=9*10^9Nm^2C^-2$

The distance

$r_{1} = \sqrt{{(- 7)}^{2} + {(5)}^{2}} = \sqrt{74} m$ $r_1=sqrt((-7)^2+(5)^2)=sqrt74m$

The distance

$r_{2} = \sqrt{{(9)}^{2} + {(- 2)}^{2}} = \sqrt{85}$ $r_2=sqrt((9)^2+(-2)^2)=sqrt(85)$

Therefore,

$U_{1} = k \frac{q_{1} q_{2}}{r_{1}}$ $U_1=k(q_1q_2)/r_1$

$U_{2} = k \frac{q_{1} q_{2}}{r_{2}}$ $U_2=k(q_1q_2)/r_2$

$DeltaU=U_2-U_1=k(q_1q_2)/r_2-k(q_1q_2)/r_1$

$=k(q_1q_2)(1/r_1-1/r_2)$

$=9*10^9*((-2)*(4))(1/sqrt74-1/sqrt(85))$

$=-0.56*10^9J$

The energy released is $=0.56*10^9J$

A charge of -2 C−2C-2 C is at the origin. How much energy would be applied to or released from a 4 C4C 4 C charge if it is moved from (-7 ,5 ) (−7,5) (-7 ,5 ) to (9 ,-2 ) (9,−2)(9 ,-2 ) ?