Two point charges q 1=+2.55nC and q 2=6.55nC are 0.100 m apart. Point A is midwa

Question

Two point charges q1=+2.55nC and q2=6.55nC are 0.100 m apart. Point A is midway between them; point B is 0.080 m from q1and 0.060 m from q2. (See (Figure 1) .) Take the electric potential to be zero at infinity.

Part A

Find the potential at point A.

Express your answer in volts to three significant figures.

Part B

Find the potential at point B.

Express your answer in volts to three significant figures.

Part C

Find the work done by the electric field on a charge of 2.50 nC that travels from point B to point A.

Express your answer in joules to two significant figures.

Explanation / Answer

step;1

Given that

charge q1=2.55 nC

charge q2=-6.55nC

step;2

now we find the electric potential at point A

the electric potential at point A

V1=9*10^9*2.55*10^-9/0.05=459 v

V2=9*10^9*-6.55*10^-9/0.05=-1179 v

the electric potential point A=>Va=-1179+459=-720v

step;3

now we find the potential difference at point B

V1=9*10^9*2.55*10^-9/0.08=286.9 v

V2=9*10^9*-6.55*10^-9/0.06=-982.5 v

the electric potential at point B=>Vb=-982.5+286.9=-696 v

step;4

now we find the work done

the work done W=(vb-Va)a=(-696+720)*2.5*10^-9=60*10^-9 J

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