Find the potential at point A. Find the potential at point B. Find the work done by the electric field on a charge of 2.80 nC that travels from point B to point A.

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Two point charges q1 = +2.30 nC and q2 = -6.40 nC are
0.100 m apart. Point A is midway between them; point
Bis 0.080 m from q1 and 0.060 m from q2. (See
(Figure 1).) Take the electric potential to be zero at
infinity.
%3D
Figure
1 of 1
B
0.080 m
+0.050 m→*-0.050 m-
91
92
0.060 m-
Transcribed Image Text:Two point charges q1 = +2.30 nC and q2 = -6.40 nC are 0.100 m apart. Point A is midway between them; point Bis 0.080 m from q1 and 0.060 m from q2. (See (Figure 1).) Take the electric potential to be zero at infinity. %3D Figure 1 of 1 B 0.080 m +0.050 m→*-0.050 m- 91 92 0.060 m-
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Find the potential at point A. Find the potential at point B. Find the work done by the electric field
on a charge of 2.80 nC that travels from point B to point A.
!!
lilı
因
Transcribed Image Text:Arial BIUA Normal text 10 | 5. Find the potential at point A. Find the potential at point B. Find the work done by the electric field on a charge of 2.80 nC that travels from point B to point A. !! lilı 因
Expert Solution
Step 1

given,q1=+2.30nCq2=-640nCr=0.100ma) potential at point A,VA=Vq1+Vq2      =kq1r1A+kq2r2Awhere r1A is distance btween charge q1 and Ar2A is distance between charge q2 and AVA=9×109(2.3×10-9C)0.05m+9×109(-6.40×10-9C)0.05m=-738Vb) potential at point B,VB=Vq1+Vq2      =kq1r1B+kq2r2Bwhere r1B is distance btween charge q1 and Br2B is distance between charge q2 and BVB=9×109(2.3×10-9C)0.08m+9×109(-6.40×10-9C)0.06m=-701.25Vc)Work done from point B to A,WBA=q(VB-VA)=2.8×10-9C(-701.25V-(-738V))=102.9nJ

 

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