A charge of - 0.15 C is moved from a position where the electric potential is 13 V to a position where the electric potential is 60 V. What is the change in potential energy of the charge associated with this change in position?
ΔPE= -Work=-qΔφ= -{(-0.15)•(13-60)} =
=-7.05 J
To calculate the change in potential energy, we can use the formula:
ΔPE = q * ΔV
where:
ΔPE is the change in potential energy,
q is the charge, and
ΔV is the change in electric potential.
Given:
q = -0.15 C (negative charge)
ΔV = 60 V - 13 V = 47 V (change in electric potential)
Substituting the given values into the formula:
ΔPE = -0.15 C * 47 V
Calculating the result:
ΔPE = -7.05 J
Therefore, the change in potential energy of the charge is -7.05 J.
To find the change in potential energy of the charge, we can use the equation:
ΔPE = q * ΔV
Where:
ΔPE is the change in potential energy,
q is the charge,
ΔV is the change in electric potential.
Given:
q = -0.15 C (negative because it represents a negative charge)
ΔV = 60 V - 13 V = 47 V (the change in potential)
Substituting the values into the equation:
ΔPE = (-0.15 C) * (47 V)
ΔPE = -7.05 J
Therefore, the change in potential energy of the charge associated with this change in position is -7.05 Joules (J).