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

Question

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

Emily Abbate · Accepted Answer

The potential energy difference between the initial and final positions of the charges will be #8.2xx10^10# #J#.

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# #m# (we have to assume the units are #m#, although we were not told) #r_"final"=sqrt(2^2+(-1)^2)=sqrt(4+1)=sqrt(5)=2.2# #m# Since both charges are positive, and they are moved closer together, electric potential energy will be added to the system. #E_"initial"=(kq_1q_2)/r=(9xx10^9xx7xx5)/5.1=6.1xx10^10# #J# #E_"final"=(kq_1q_2)/r=(9xx10^9xx7xx5)/2.2=1.2xx10^11# #J# #E_"change"=E_"final"-E_"initial"# #=1.4xx10^11-6.1xx10^10=8.2xx10^10# #J#

Olivia Askew · Answer

undefined The energy applied to or released from a charge when it is moved in an electric field can be calculated using the formula:

$$W = q \cdot V$$

Where $W$ is the work done, $q$ is the charge, and $V$ is the electric potential difference.

First, we need to calculate the electric potential difference $V$ between the initial and final positions of the charge. This can be found using the formula:

$$V = k \cdot \frac{q}{r}$$

Where $k$ is Coulomb's constant ($8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$), $q$ is the charge, and $r$ is the distance between the charges.

The initial position of the 5 C charge is at (-5, 1) and the final position is at (2, -1). The distance between these points can be calculated using the distance formula:

$$r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Plugging in the values, we get:

$$r = \sqrt{(2 - (-5))^2 + (-1 - 1)^2} = \sqrt{49 + 4} = \sqrt{53}$$

Now, we can calculate the electric potential difference $V$:

$$V = (8.99 \times 10^9) \cdot \frac{7}{\sqrt{53}}$$

$$V ≈ 1.65 \times 10^9 \, \text{V}$$

Now, we can calculate the work done ($W$):

$$W = (5 \, \text{C}) \cdot (1.65 \times 10^9 \, \text{V})$$

$$W ≈ 8.25 \times 10^9 \, \text{J}$$

Therefore, approximately $8.25 \times 10^9 \, \text{J}$ of energy would be applied to or released from the 5 C charge when it is moved from (-5, 1) to (2, -1).