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 # (6, -1 ) # to #(-9 ,-6 ) #?

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 # (6,  -1  ) # to #(-9 ,-6  )  #?

Layla Ahmed · Accepted Answer

The energy released is #=22.7*10^9J#

Potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=7C#

The charge #q_2=5C#

The Coulomb's constant is #k=9*10^9Nm^2C^-2#

The separation

#r_1=sqrt((6^2+(-1)^2))=sqrt37m#

The separation

#r_2=sqrt((-9)^2+(-6)^2)=sqrt(117)#

Consequently,

#U_1=k(q_1q_2)/r_1#

#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_2-1/r_1)#

#=9*10^9*((7)*(5))(1/sqrt117-1/sqrt(37))#

#=-22.7*10^9J#

The energy released is #=22.7*10^9J#

Adam Arnold · Answer

undefined The electric potential energy between two point charges is given by the equation $ U = \frac{kQq}{r} $, where $ k $ is Coulomb's constant ($ 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 $), $ Q $ and $ q $ are the magnitudes of the charges, and $ r $ is the distance between them.

Using the given coordinates, we can find the distance between the charges using the distance formula: $ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $.

Substituting the given values into the distance formula, we find $ r = \sqrt{(-9 - 6)^2 + (-6 - (-1))^2} = \sqrt{225} = 15 $ m.

Then, substituting the values of the charges and the distance into the electric potential energy formula, we get $ U = \frac{(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2)(7 \, \text{C})(5 \, \text{C})}{15 \, \text{m}} = 2.996 \times 10^9 \, \text{J} $.

Therefore, the energy applied to or released from the 5 C charge when it is moved from (6, -1) to (-9, -6) is $ 2.996 \times 10^9 \, \text{J} $.