A point charge q_1 = 4.00 nC is placed at the origin, and a second point charge q_2 = -3.00 nC, placed on the x-axis at x = + 20.0 cm. A third point charge q_3 = 2.00 nC is placed on the X-axis between q_1, and q_2. (Take as zero the potential energy of the three charges when they are infinitely far apart). (a) What is the potential energy of the system of the three charges if q_3 is placed at x= + 10.0 cm?
(b) Where should q_3 be placed to make the potential energy of the system equal to zero?
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Answer:
Here three charges q
1
=+2nC, q
2
=−2nC and q
0
=1μC are at (0,2), (0,−2) and (2,0) respectively. The charges are at vertices of a right angle triangle, and charge q
0
is at the right angle vertex.
Now force between q
1
and q
0
which is repulsive in nature is given by
F=
(2
2
×10
−2
)
2
9×10
9
×2×10
−9
×10
−6
F=
8
18×10
−2
F=2.25×10
−2
N
Similarly force between q
2
and q
0
is also same but attractive in nature and perpendicular to the direction of first force.
Resultant force
F
R
=
2.25
2
+2.25
2
×10
−2
F
R
=3.18×10
−2
N
F
R
=0.0318N in downward vertical direction i.e. negative j direction.
Thus F=−0.0318jN
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