An electric force moves a charge of +1.85 10-4 C from point A to point B and performs 6.70 10-3 J of work on the charge.
(a) What is the difference (EPEA - EPEB) between the electric potential energies of the charge at the two points?
To find the difference in electric potential energies between point A and point B, we need to use the equation:
∆EPE = EPEA - EPEB
where ∆EPE is the difference in electric potential energy, EPEA is the electric potential energy at point A, and EPEB is the electric potential energy at point B.
The electric potential energy is given by the equation:
EPE = qV
where EPE is the electric potential energy, q is the charge, and V is the electric potential (also known as voltage).
In this case, the charge is +1.85 * 10^-4 C.
To calculate the electric potential at each point, we need to use the equation:
V = W/q
where V is the electric potential, W is the work done on the charge, and q is the charge.
In this case, the work done on the charge is 6.70 * 10^-3 J.
Now, let's substitute the values into the equations to find the electric potential at each point:
Electric potential at point A (VA) = W/q = 6.70 * 10^-3 J / (1.85 * 10^-4 C)
Electric potential at point B (VB) = W/q = 6.70 * 10^-3 J / (1.85 * 10^-4 C)
Now, we can find the difference in electric potential energies:
∆EPE = EPEA - EPEB = qVA - qVB
Substitute the calculated values of VA and VB into the equation:
∆EPE = (1.85 * 10^-4 C) * (6.70 * 10^-3 J / (1.85 * 10^-4 C)) - (1.85 * 10^-4 C) * (6.70 * 10^-3 J / (1.85 * 10^-4 C))
Simplifying the expression:
∆EPE = 6.70 * 10^-3 J - 6.70 * 10^-3 J
∆EPE = 0 J
Therefore, the difference in electric potential energies between point A and point B is 0 J.
ΔPE=A= q •Δφ
Δφ= A/q=6.7•10⁻³/1.85•10⁻⁴=36 V