A -5.50 nC point charge is on the x axis at x = 1.35 m. A second point charge is
ID: 2226811 • Letter: A
Question
A -5.50 nC point charge is on the x axis at x = 1.35 m. A second point charge is on the x axis at x = -0.650 m.
a) What must be the charge Q for the resultant electric field at the origin to be 48.5 N/C in the +xdirection?
A -5.50 nC point charge is on the x axis at x = 1.35 m. A second point charge is on the x axis at x = -0.650 m. a) What must be the charge Q for the resultant electric field at the origin to be 48.5 N/C in the +xdirection? b)What must be the charge -x direction? N/C in the Q for the resultant electric field at the origin to be48.5Explanation / Answer
E1=kq1/x2
k=8.988*109
x-distance from origin of charge 1.
lly, E2=kq2/x2
x-distance from origin of charge 2.
resultant E.F(E)= E1+E2
A)
Resultant is in positive X drn
=> E is positive.
E1=8.988*109(-5.50)*10-9/1.352
E2=8.988*109(Q)*10-9/0.652
=>
[8.988*109(-5.50)*10-9/1.352 ]+[8.988*109(Q)*10-9/0.652] =+48.5
Q/0.652-5.50/1.352=48.5/8.988
Q=0.652(5.396+3.0178)
=0.4225(8.414)
=3.5549
therefore, Q =3.5549nC
B)
Resultant is in negative X drn
=> E is negative.
E1=8.988*109(-5.50)*10-9/1.352
E2=8.988*109(Q)*10-9/0.652
=>
[8.988*109(-5.50)*10-9/1.352 ]+[8.988*109(Q)*10-9/0.652] =-48.5
Q/0.652-5.50/1.352=-48.5/8.988
Q=0.652(-5.396+3.0178)
=0.4225(-2.378)
=-1.0047
therefore, Q =-1.0047nC
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