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Question

Two long parallel plates are separated by a distance of 15.0 cm. The bottom plate is kept at a potential of −250 V and the top at

+250V+ 250 \mathrm { V }

. A charge of

2.00μC- 2.00 \mu \mathrm { C }

is placed at a point 3.00 cm from the bottom plate. a. Find the electric potential energy of the charge. The charge is then moved vertically up to a point 3.00 cm from the top plate. b. What is the electrical potential energy of the charge now? c. How much work was done on the charge?

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.a)\textbf{.a)} The potential at a distance (s)(s) from the bottom plate can be written as follows

V=Vbottom+E×sV=V_{\text{bottom}}+E\times s

Where (Vbottom)(V_{\text{bottom}}) is the electric potential of the bottom plate and (E)(E) is the electric field between the two plates, and we know that the electric field between the two plates is (ΔV/Δr)(\Delta V/\Delta r), where (Δr=15(\Delta r=15 cm). Thus, the electric potential at a distance (s)(s) from the bottom plate is

V=250 V+250 V(250 V)0.15 m×(0.03 m)V=-250 \mathrm{~ V}+\frac{250\mathrm{~ V} -(-250 \mathrm{~ V})}{0.15 \mathrm{~ m}} \times (0.03 \mathrm{~ m})

V=150 VV=-150 \mathrm{~ V}

now that we have the potential at the position of the charge, we can calculate its electric potential energy as follows

Ep=qV=(2×106 C)×(150 V)=3×104 JE_{p}=qV=(- 2 \times 10^{-6} \mathrm{~ C})\times (-150 \mathrm{~ V})=3\times 10^{-4} \mathrm{~ J}

.b)\textbf{.b)} The charge is now at a distance of (12 cm) from the bottom plate, so the potential at this point can be determined as follows

V=250 V+250 V(250 V)0.15 m×(0.12 m)V=-250 \mathrm{~ V}+\frac{250\mathrm{~ V} -(-250 \mathrm{~ V})}{0.15 \mathrm{~ m}} \times (0.12 \mathrm{~ m})

V=150 VV=150 \mathrm{~ V}

and the electric potential energy is therefore

Ep=qV=(2×106 C)×(150 V)=3×104 JE_{p}=qV=(- 2 \times 10^{-6} \mathrm{~ C})\times (150 \mathrm{~ V})=-3\times 10^{-4} \mathrm{~ J}

.c)\textbf{.c)} The work equals the change in the potential energy

W=ΔEp=3×104 J3×104 J=6×104 JW=\Delta E_{p}=-3\times 10^{-4} \mathrm{~ J}-3\times 10^{-4} \mathrm{~ J}=-6\times 10^{-4} \mathrm{~ J}

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