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 # (-7 ,1 ) # to #(4 ,-6 ) #?

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 # (-7   ,1  ) # to #(4  ,-6  )  #?

Julia Aguilar · Accepted Answer

#197# #MJ# of energy is released.

Electrostatic Potential Energy: The electrostatic potential energy of a charge pair #Q# and #q# separated by a distance #r# is : #U(r) = -k(Qq)/r; \qquad k \equiv 1/(4\pi\epsilon_0) = 8.99 imes10^9 (N.m^2)/C^2# Assuming that the units of the coordinates are in meters. #r_i = \sqrt{(-7m)^2+(1m)^2} = \sqrt{50}# #m = 7.071# #m# #r_f = \sqrt{(4m)^2+(-6m)^2} = 7.211# #m# #\Delta U = U_f - U_i = -kQq(1/r_f - 1/r_i)# #= -(8.99 imes10^9 (Nm^2)/C^2)(-2C)(4C)(1/(7.211m) - 1/(7.071m))# #= - 197# #MJ# A negative change in potential energy indicates a decrease in potential energy, which releases the energy.

Isabella Alvarez · Answer

undefined The energy applied to or released from a 4 C charge when moved from (-7, 1) to (4, -6) due to the presence of a -2 C charge at the origin can be calculated using the formula for electric potential energy:

U = k * (q1 * q2) / r

Where:
U is the electric potential energy
k is Coulomb's constant (8.99 × 10^9 N m²/C²)
q1 and q2 are the charges (in coulombs)
r is the distance between the charges (in meters)

First, we need to calculate the distance between the two charges using the distance formula:

Distance = √((x2 - x1)² + (y2 - y1)²)

Plugging in the coordinates, we get:

Distance = √((4 - (-7))² + ((-6) - 1)²) = √(11² + (-7)²) = √(121 + 49) = √170

Now, we can calculate the energy:

U = (8.99 × 10^9 N m²/C²) * ((-2 C) * (4 C)) / (√170 m)
U ≈ -9.59 × 10^8 J

The negative sign indicates that energy is released from the 4 C charge as it moves from (-7, 1) to (4, -6) in the presence of the -2 C charge at the origin.