# A charge of -6 C is at the origin. How much energy would be applied to or released from a  9 C charge if it is moved from  (5 ,-2 )  to (2 ,-1 ) ?

Jan 13, 2018

(127.09)*(10^9) Joule amount of energy has to be taken out from the charge.

#### Explanation:

Initial potential energy of the charge is,
Ei = (9*10^9)*{(-6)*9}/(√29) [r=distance between them = √{(5)^2+(-2)^2}]
And final potential energy=
Ef = (9*10^9)*{(-6)*9}/(√5) [r1=distance between them =√{(2)^2+(-1)^2}
Hence change in energy =
Ef-Ei
= (9*10^9){(-6)*9}{1/(√5)-1/(√29)}

Or,-$\left(127.09\right) \cdot \left({10}^{9}\right)$ Joule
Negative sign means energy has to be taken out from the charge

Jan 13, 2018

The energy applied is $= 127.1 \cdot {10}^{9} J$

#### Explanation:

The potential energy is

$U = k \frac{{q}_{1} {q}_{2}}{r}$

The charge ${q}_{1} = - 6 C$

The charge ${q}_{2} = 9 C$

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

The distance

${r}_{1} = \sqrt{{\left(5\right)}^{2} + {\left(- 2\right)}^{2}} = \sqrt{29} m$

The distance

${r}_{2} = \sqrt{{\left(2\right)}^{2} + {\left(1\right)}^{2}} = \sqrt{5}$

Therefore,

${U}_{1} = k \frac{{q}_{1} {q}_{2}}{r} _ 1$

${U}_{2} = k \frac{{q}_{1} {q}_{2}}{r} _ 2$

$\Delta U = {U}_{2} - {U}_{1} = k \frac{{q}_{1} {q}_{2}}{r} _ 2 - k \frac{{q}_{1} {q}_{2}}{r} _ 1$

$= k \left({q}_{1} {q}_{2}\right) \left(\frac{1}{r} _ 1 - \frac{1}{r} _ 2\right)$

$= 9 \cdot {10}^{9} \cdot \left(\left(- 6\right) \cdot \left(9\right)\right) \left(\frac{1}{\sqrt{29}} - \frac{1}{\sqrt{5}}\right)$

$= 127.1 \cdot {10}^{9} J$