Question

A charge 5.05 nC is placed at the origin of
an*xy*-coordinate system, and a charge -1.97 nC is placed on
the positive *x*-axis at *x* = 3.99 cm . A third
particle, of charge 6.03 nC is now placed at the point *x* =
3.99 cm , *y* = 3.00 cm .

(1)Find the *x*-component of the total force exerted on
the third charge by the other two.

(2)Find the *y*-component of the total force exerted on
the third charge by the other two.

(3)Find the magnitude of the total force acting on the third charge.

(4)Find the direction of the total force acting on the third charge.

Answer #1

Given that :

charge on q_{1} = 5.5 x 10^{-9}
C
at x = 0

charge on q_{2} = -1.97 x 10^{-9}
C
at x = 3.99 cm

charge on q_{3} = 6.03 x 10^{-9}
C
at x = 3.99 cm & y = 3 cm

Distance of q_{3} from q_{1}, r_{31} =
_{}(3.99
cm)^{2} + (3 cm)^{2} = 4.99 x 10^{-2} m

Distance of q_{3} from q_{2}, r_{32} = 3
x 10^{-2} m

The force exerted on q_{3} due to q_{1} which is
given as :

F_{31} = k q_{3} q_{1} /
(r_{31})^{2}
{ eq.1 }

F_{31} = (9 x 10^{9}
Nm^{2}/C^{2}) (6.03 x 10^{-9} C) (5.5 x
10^{-9} C) / (4.99 x 10^{-2} m)^{2}

F_{31} = 11 x 10^{-5}
N
(along a positive axis)

using an identity, tan = (3 cm) (3.99 cm)

=
tan^{-1} (0.7518)

= 36.94 degree

component along x axis, F_{X} = F Cos
{ eq.2 }

inserting the values in eq.2,

F_{X} = (11 x 10^{-5} N) Cos
(36.94^{0})

F_{X} = 7.97 x 10^{-5} N (i)

component along y axis, F_{Y} = F Sin
{ eq.3 }

inserting the values in eq.3,

F_{y} = (11 x 10^{-5} N) Sin
(36.94^{0})

F_{y} = 6.6 x 10^{-5} N (j)

Now, we have F_{31} = F_{X} +
F_{Y} { eq.4 }

inserting the values in eq.4,

F_{31} = [7.97 x 10^{-5} N (i) + 6.6 x
10^{-5} N (j)]

The force exerted on q_{3} due to q_{2} which is
given as :

F_{32} = k q_{3} q_{2} /
(r_{32})^{2}
{ eq.5 }

F_{32} = (9 x 10^{9}
Nm^{2}/C^{2}) (6.03 x 10^{-9} C) (-1.97 x
10^{-9} C) / (3 x 10^{-2} m)^{2}

F_{32} = 11.8 x 10^{-5} N (-j)
(along a negative axis)

Total force, F_{total} = F_{31} +
F_{32}
{ eq.6 }

then F_{total} =
[7.97 x 10^{-5} N (i) + 6.6 x 10^{-5} N (j)] +
[11.8 x 10^{-5} N (-j)]

F_{total} = 7.97 x 10^{-5} N (i) - 5.2 x
10^{-5} N (-j)

(a) the *x*-component of the total force exerted on the
third charge by the other two which is given as ::

**F _{x} = 7.97 x 10^{-5}
N **
(in +ve x-direction)

(b) the *y*-component of the total force exerted on the
third charge by the other two which is given as :

**F _{y} = 5.2 x 10^{-5} N**
(in -ve y-direction)

(c) magnitude of the total force acting on the third charge will be given as :

F_{net} = _{}(7.97
x 10^{-5} N)^{2} + (5.2 x 10^{-5}
N)^{2}

F_{net} = _{}90.56
x 10^{-10}

**F _{net} = 9.51 x 10^{-5} N**

(4) direction of the total force acting on the third charge will be given as :

=
tan^{-1} [(5.2 x 10^{-5} N) / (7.97 x
10^{-5} N)]

=
tan^{-1} (0.6524)

** = 33.12
degree**

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