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Q1: Two positive charges Q1 and Q2 are held together by a string of length d=10[cm]. The first charge is Q1=5[C] and the tension in the string is equal to 1.348132768×1013[N]. Find the value of Q2.
A1: Q2=Fed25k=+3[C].
Q2: Suppose that a charge Q1=+5[C] is placed at the origin and a charge Q2=−3[C] is placed at distance d[m] along the x axis. There exists a point x which lies somewhere between the charges Q1 and Q2 where the electric potential (V) is equal to zero. Find the value of x.
Hint: V\tiny tot(x)=V1(x)+V2(x)=kQ1x+kQ2d−x.
Ans: x=58d.
Q3: Suppose that we have same situation as in the above question:
a charge Q1=+5[C] is placed at the origin and a charge Q2=−3[C] is placed at distance (d,0)[m].
Find the find the point x where the strength of the electric field is equal to zero.
A3: x=(5+√15)2d≈4.4365d.
Q4: A charge Q1 is placed at the origin of the coordinate system,
and a second charge Q2 is placed at distance d[m] along the x axis.
a) Find the magnitude of the electric force between Q1 and Q2.
b) Find the electric potential energy stored in this configuration.
c) Find the electric field →E at a point P which has x coordinate
x=5d.
This point P is situated at a distance 4d from the charge Q2.
d) Find the electric potential (V) at the point P.
e) If a third charge Q3 is placed at the point P,
what will be the electric potential energy of the the configuration of the charges.
Use your answers from part b) and d) to answer this question.
A4: a) Fe=kQ1Q2d2[N],
b) U12=kQ1Q2d[J],
c) →E\tiny tot(P)=→E1(P)+→E2(P)=kQ1(5d)2ˆı+kQ2(4d)2ˆı [N/C].
d) V\tiny 12(P)=V1(P)+V2(P)=kQ15d+kQ24d [V].
e) U\tiny tot=U12+U13+U23=kQ1Q2d+Q3V\tiny 12(P) [J].