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

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

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Answer 1 · 2017-06-24T13:27:17

The potential energy difference between the initial and final positions of the charges will be $8.2 \times 10^{10}$ $8.2xx10^10$ $J$ $J$ .

Explanation:

We care only about the distance between the two charges. We can use Pythagoras' theorem to calculate the distances and the change:

$r_{initial} = \sqrt{(- 5^{2}) + 1^{2}} = \sqrt{25 + 1} = \sqrt{26} = 5.1$ $r_"initial"=sqrt((-5^2)+1^2)=sqrt(25+1)=sqrt(26)=5.1$ $m$ $m$ (we have to assume the units are $m$ $m$ , although we were not told)

$r_{final} = \sqrt{2^{2} + {(- 1)}^{2}} = \sqrt{4 + 1} = \sqrt{5} = 2.2$ $r_"final"=sqrt(2^2+(-1)^2)=sqrt(4+1)=sqrt(5)=2.2$ $m$ $m$

Since both charges are positive, and they are moved closer together, electric potential energy will be added to the system.

$E_{initial} = \frac{k q_{1} q_{2}}{r} = \frac{9 \times 10^{9} \times 7 \times 5}{5.1} = 6.1 \times 10^{10}$ $E_"initial"=(kq_1q_2)/r=(9xx10^9xx7xx5)/5.1=6.1xx10^10$ $J$ $J$

$E_{final} = \frac{k q_{1} q_{2}}{r} = \frac{9 \times 10^{9} \times 7 \times 5}{2.2} = 1.2 \times 10^{11}$ $E_"final"=(kq_1q_2)/r=(9xx10^9xx7xx5)/2.2=1.2xx10^11$ $J$ $J$

$E_{change} = E_{final} - E_{initial}$ $E_"change"=E_"final"-E_"initial"$
$= 1.4 \times 10^{11} - 6.1 \times 10^{10} = 8.2 \times 10^{10}$ $=1.4xx10^11-6.1xx10^10=8.2xx10^10$ $J$ $J$

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

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Explanation:

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Explanation:

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