A charge of $7 C$ $7 C$ is at the origin. How much energy would be applied to or released from a $9 C$ $9 C$ charge if it is moved from $(5, - 2)$ $(5 ,-2 )$ to $(2, - 1)$ $(2 ,-1 )$ ?

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Answer 1 · 2018-02-02T13:36:55

The energy applied is $= 148.3 \cdot 10^{9} J$ $=148.3*10^9J$

The potential energy is

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

The charge $q_{1} = 7 C$ $q_1=7C$

The charge $q_{2} = 9 C$ $q_2=9C$

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

The distance

$r_{1} = \sqrt{{(5)}^{2} + {(- 2)}^{2}} = \sqrt{29} m$ $r_1=sqrt((5)^2+(-2)^2)=sqrt29m$

The distance

$r_{2} = \sqrt{{(2)}^{2} + {(- 1)}^{2}} = \sqrt{5}$ $r_2=sqrt((2)^2+(-1)^2)=sqrt(5)$

Therefore,

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

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

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

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

$=9*10^9*((7)*(9))(1/sqrt29-1/sqrt(5))$

$=-148.3*10^9J$

The energy needed is $=148.3*10^9J$

A charge of 7 C7C7 C is at the origin. How much energy would be applied to or released from a 9 C9C 9 C charge if it is moved from (5 ,-2 ) (5,−2) (5 ,-2 ) to (2 ,-1 ) (2,−1)(2 ,-1 ) ?