A charge of #-2 C# is at the origin. How much energy would be applied to or released from a # 4 C# charge if it is moved from # (-5 ,3 ) # to #(3 ,-4 ) #?

Question

A charge of #-2  C# is at the origin. How much energy would be applied to or released from a # 4  C# charge if it is moved from # (-5  ,3   ) # to #(3 ,-4   )  #?

Emily Abbate · Accepted Answer

The energy released is #=2.052*10^9J#

Potential energy is

#U=k(q_1q_2)/r#

The charge #q_1=-2C#

The charge #q_2=4C#

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

The separation

#r_1=sqrt((-5)^2+(3)^2)=sqrt(34)m#

The separation

#r_2=sqrt((3)^2+(-4)^2)=sqrt(25)=5m#

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_1-1/r_2)#

#=9*10^9*((-2)*(4))(1/5-1/sqrt(34))#

#=-2.052*10^9J#

Layla Ahmed · Answer

undefined To calculate the change in energy, you can use the formula:

ΔU = k * (q1 * q2) / r1 - (q1 * q2) / r2

Where:
- ΔU is the change in energy
- k is Coulomb's constant (8.99 × 10^9 N*m^2/C^2)
- q1 and q2 are the charges (-2 C and 4 C, respectively)
- r1 and r2 are the distances between the charges at the initial and final positions, respectively.

First, calculate the distances:
- Initial distance: √((-5 - 0)^2 + (3 - 0)^2) = √(25 + 9) = √34
- Final distance: √((3 - 0)^2 + (-4 - 0)^2) = √(9 + 16) = √25 = 5

Substitute the values into the formula:
ΔU = (8.99 × 10^9) * ((-2 * 4) / √34 - (-2 * 4) / 5)
   ≈ (8.99 × 10^9) * ((-8 / √34) - (-8 / 5))
   ≈ (8.99 × 10^9) * (-8 / √34 + 8 / 5)
   ≈ (8.99 × 10^9) * (-8 / √34 + 8 / 5)
   ≈ (8.99 × 10^9) * (-2.401 - 1.28)
   ≈ (8.99 × 10^9) * (-3.681)
   ≈ -3.31179 × 10^10 J

Therefore, approximately -3.31 × 10^10 joules of energy would be applied to the 4 C charge as it moves from (-5, 3) to (3, -4).