A charge of $- 5 C$ $-5 C$ 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 $(- 5, 3)$ $(-5, 3 )$ to $(2, - 7)$ $(2 ,-7 )$ ?

Question

A charge of $- 5 C$ $-5 C$ 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 $(- 5, 3)$ $(-5, 3 )$ to $(2, - 7)$ $(2 ,-7 )$ ?

1 Answer

Impact of this question

1430 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-30T12:13:20

$W = 6.12 \cdot 10^{9} Joules$ $W=6.12*10^9" Joules"$
$since The Potential of the point B is greater than the potential$ $"since The Potential of the point B is greater than the potential"$
$of the point of A, work is done by an external electrical force.$ $"of the point of A, work is done by an external electrical force. "$

Explanation:

enter image source here

$since The electrical force is conservative,it is not important$ $"Since The electrical force is conservative,it is not important"$
$that the charge have gone from A to B by which path.(f or c).$ $"that the charge have gone from A to B by which path.(f or c)."$ $Work doing both situation is the same.$ $"Work doing both situation is the same."$

$since a charge of 4C moves in the electric field of the charge -5C,$ $"Since a charge of 4C moves in the electric field of the charge -5C,"$
$we should find the potentials of the points A and B.$ $"we should find the potentials of the points A and B."$

$V_{a} = - k \cdot \frac{5}{r_{a}}, V_{b} = - k \frac{5}{r_{b}}$ $V_a=-k*5/(r_a)" , "V_b=-k5/(r_b)$

$r_{a} = \sqrt{{(- 5)}^{2} + 3^{2}} = \sqrt{25 + 9} = \sqrt{34} = 5.83$ $r_a=sqrt((-5)^2+3^2)=sqrt(25+9)=sqrt(34)=5.83$

$r_{b} = \sqrt{2^{2} + {(- 7)}^{2}} = \sqrt{4 + 49} = \sqrt{53} = 7.28$ $r_b=sqrt(2^2+(-7)^2)=sqrt(4+49)=sqrt(53)=7.28$

$W = Q (V_{b} - V_{a})$ $W=Q(V_b-V_a)$

$W = 4 (- k \cdot \frac{5}{r_{b}} + k \cdot \frac{5}{r_{a}})$ $W=4(-k*5/r_b+k*5/r_a)$

$W = 4 \cdot 5 k (\frac{1}{r_{a}} - \frac{1}{r_{b}})$ $W=4*5k(1/r_a-1/r_b)$

$W = 20 k (\frac{r_{b} - r_{a}}{r_{a} \cdot r_{b}})$ $W=20k((r_b-r_a)/(r_a*r_b))$

$W = 20 k (\frac{7.28 - 5.83}{7.28 \cdot 5.83})$ $W=20k((7.28-5.83)/(7.28*5.83))$

$k = 9 \cdot 10^{9}$ $k=9*10^9$

$W = 20 k (0.034)$ $W=20k(0.034)$

$W = 0.68 \cdot 9 \cdot 10^{9}$ $W=0.68*9*10^9$

$W = 6.12 \cdot 10^{9} Joules$ $W=6.12*10^9" Joules"$

Science

Math

Humanities

... and beyond

A charge of $- 5 C$ $-5 C$ 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 $(- 5, 3)$ $(-5, 3 )$ to $(2, - 7)$ $(2 ,-7 )$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A charge of -5 C−5C-5 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 (-5, 3 ) (−5,3) (-5, 3 ) to (2 ,-7 ) (2,−7)(2 ,-7 ) ?

1 Answer

Explanation:

Related questions

Impact of this question

A charge of $- 5 C$ $-5 C$ 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 $(- 5, 3)$ $(-5, 3 )$ to $(2, - 7)$ $(2 ,-7 )$ ?