Question

Particle A of charge 3.15  10^-4 C is at the origin, particle B of charge -5.94  10^-4 C is at (4.00 m, 0), and particle C of charge 1.05  10^-4 C is at (0, 3.00 m). We wish to find the net electric force on C.

(a) What is the x component of the electric force exerted by A on C?
N

(b) What is the y component of the force exerted by A on C?
N

(c) Find the magnitude of the force exerted by B on C.
N

(d) Calculate the x component of the force exerted by B on C.
N

(e) Calculate the y component of the force exerted by B on C.
N

(f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C.
N

(g) Similarly, find the y component of the resultant force vector acting on C.
N

(h) Find the magnitude and direction of the resultant electric force acting on C.

magnitude	N
direction	° counterclockwise from the +x-axis

Answer 1

The electric force exerted by a charged particle on other is given by

F = kq₁q_​​​2/r^{​​​​2}

^{where q_{​​​​​​1 and} q_{​​​​​2 are the charge particle , r is the distance between them and k is constant.}}

The charge of paticle A is 3.15 x 10^-4 C at position (0,0), particle B is -5.94 x10^-4 C at position (4m, 0) and particle C is 1.05 x 10^-4C at position (0,3m).

The distance between C and B is

r_{​​​​​​cb^{​​​​2} = 4² + 3²}

r _{​​​cb = 5m}

a. The distance between A snd C in x component is 0. So, the force exerted zero. ( Mathematically its infinite but we take physically zero value)

b. The distance between A and C in y component is 3m. So. The force exerted by A on C is

F = kq_aq_​​​c/r^{​​​​​​2}

= ( 9 x 10 ^9 N m^{​​​​​​2} C^{​​​​​​-2} )(3.15 x 10^-4 C) (1.05 x 10 ^-4 C)/ 3²  

= 33.075 N

c. The magnitude of force exerted on C by B is

F = kq_bq_​​​c/r^{​​​​​​2}

= (9 x 10^9)(-5.94 × 10^-4)(1.05 × 10^-4) / 5²

=-22.4532 N

d. The distance between B and C in x component is 4m. So the force exerted by B on C is

F = kq_bq_​​​c/r^{​​​​​​2}

= (9 x 10^9)(-5.94 x 10^-4)(1.05 x 10^-4) / 4²

=-35.0831 N

e. The distance between B and C in y component is 3m. So, the force exerted on C by B is

F = kq_bq_​​​c/r^{​​​​​​2}

= (9 x 10^9)(-5.94 x 10^-4)(1.05 x 10^-4) / 9

= -62.37 N

f. The magnitude of electric force in x component

F_{​​​​​​X} = 0 + (-35.083) = -35.083 N

g. The magnitude of electric force in y component

F_{​​​​​​Y} = 33.075 + (-62.37) = -29.295 N

h. The resultant force acting on C is

F = √ Fx^{​​​​​2} + Fy^{​​​​​2}

=√{(-35.083)² + (-29.295)²}

= 45.7057 N

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