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What is the electric potential at points A, B, and C in FIGURE EX 25.27?

b. What are the potential differences

ΔVAB = VB - VA and

ΔVCB = VB - VC?

Short Answer

Expert verified

1. The potential energy at the A, B, C points are 1.80 k.V, 1.80 k.V and 0.90 k.V

2. The potential energy of electrical at the A point is -2.8×10-16J

Step by step solution

01

Definition of electrical potential 

The value of q in C unit is2.0×10-19C

Transforming the value of r at A point from cm to m is 0.01m

Transforming the value of r at B point from cm to m is 0.01m

Transforming the value of r at C point from cm to m is 0.02m

VA=14πε0qrA14πε0=9.0×109N.m2/C2q=2.0×10-9CrA=0.01mVA=9.0×109N.m2/C22.0×10-9C0.01m=1.80kVVB=14πε0qrBVB=9.0×109N.m2/C22.0×10-9C0.01m=1.80kVVC=9.0×109N.m2/C22.0×10-9C0.02m=0.90kV[rC=0.02m]


02

Step 2:Definition of the potential energy of  electrical at the A point 

The formula of the potential energy of electrical is

EA=14πε0qqerA14πε0=9.0×109N.m2/C2q=2.0×10-9CrA=0.01mqe=-1.6×10-19CEA=9.0×109N.m2/C22.0×10-9C×-1.6×10-19C0.01m=-2.8×10-16JEB=14πε0qqerB=9.0×109N.m2/C22.0×10-9C×-1.6×10-19C0.01m=-2.8×10-16JEC=9.0×109N.m2/C22.0×10-9C×-1.6×10-19C0.02m=-1.4×10-16J[whererc=0.02m]

Hence the potential Difference is;

∆VAB=VB-VA=(1.80-1.80)kV=0

and the potential difference of B and C is

∆VBC=VC-vB=(-0.90-1.80)kV=-0.90kV

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