Particle A of charge 3.30 times 10^-4 C is at the origin, particle B of charge -

Question

Particle A of charge 3.30 times 10^-4 C is at the origin, particle B of charge -6.24 times 10^-4 C is at (4.00 m, 0), and particle C of charge 1.02 times 10^-4 C is at (0, 3.00 m). We wish to find the net electric force on C. What is the x component of the electric force exerted by A on C? What is the y component of the force exerted by A on C? Find the magnitude of the force exerted by B on C. Calculate the x component of the force exerted by B on C. Calculate the y component of the force exerted by B on C. Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C. Similarly, find the y component of the resultant force vector acting on C. Find the magnitude and direction of the resultant electric force acting on C. magnitude direction counter clockwise from the + x- axis

Explanation / Answer

(a) A and C are both positive ... so they repel each other

The force exerted by A on C [F(AC)] is along the y-axis so has no x-component

so x-component of F(AC) = 0 N

(b) we know that electric force F = kQ1Q2 / r2

y-component of F(AC) = [9 x 10^9 x 3.30 x 10^(-4) x 1.02 x 10^(-4)] / 3.00² = 33.66 N

(c) magnitude of F(BC) = [9 x 10^9 x -6.24 x 10^(-4) x 1.02 x 10^(-4)] / (4² + 3²) = -22.91 N ... [negative just means it's an attractive force]

(d) angle ACB = arctan (3.00 / 4.00) = 36.86°

so angle F(BC) makes with the positive x-axis is 270 + 36.86° = 306.86°

so x-component of F(BC) = -22.91 sin 306.86° = 18.32 N

(e) y-component of F(BC) =-22.91 cos 306.86° = -13.74 N (minus shows that it is toward left )

(f)

Sum the two x components from part (a) and (d) = the resultant x component of the electric force acting along x -axis

x-component of the resultant electric force on C = 0 + 18.32 = 18.32 N

(g)

Sum the two y components from part (a) and (d) = the resultant y component of the electric force acting along y -axis

y-component of the resultant electric force on C = 33.66 + -13.74 = 19.92 N

(h) magnitude of the resultant electric force on C = [18.32² + 19.92²] = 27.06 N

Direction of the resultant electric force on C = arctan (18.32 / 19.92) = 42.60° right from the y-axis ...

so that is 90 - 42.60° = 47.39° counterclockwise from the +x-axis

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