Two identical balls, each having a charge of 2 * 10 ^-7 C and mass of 100 g, are suspended from a common point

by two insulating strings each 50 cm long. The balls are held at a separation 5 cm apart and then released. Find (a) the electric force on one of the charged balls (b) the components of the resultant force on it along and perpendicular to the string (c) the tension in the string (d) the acceleration of one of the balls. Answers are to be obtained only for the instant just after the release.





Please help me find the solution to this problem along with the diagram. You can send me a diagram ( paint brush) as an attachment to my email which is buttercup_72002@yahoo.com


I'll me very grateful to the one who gets me the solution. Thank you so much :)

Two identical balls, each having a charge of 2 * 10 ^-7 C and mass of 100 g, are suspended from a common point
Sorry i cant make a sketch. But you can easily do it by yourself:


The balls and and the suspension point form an isosceles triangle with the side length s = 50cm = 0.5m and the base line b = 5cm = 0.05m





The angle β between the strings at the suspension point can evaluated by basic trigonometry. The angle bisector of β cuts the isosceles triangle into two right triangles with


hypotenuse s and b/2 as cathetus adjacent to β/2. Hence


sin(β/2) = b/2 / s = 0.05


→


β/2 = 2.87°


(you need this angle later)








(a)


The electrostatic force on the ball is given by Coulomb's law:


F_e = k · q₁ · q₂ / r²


= 8.998·10^9 Nm²/C²· · (2·10^(-7)C)² / (0.05m)² = 0.144N


This force is a repulsive force acting along the base of the triangle b.





(b)


The second force acting on the balls is gravity:


F_g = m · g = 0.1kg · 9,81= 0.981N


It acts downwards and perpendicular to the electrical force.





To split the two forces into components with respect to the direction of the string, drop a perpendicular to the direction line of the string. The two components and the force vector will form a right triangle with the components as legs. These are are given by


along string: F_s = F·cosα


perpendicular to string: F_p = F·sinα


where α is the angle between the force and the string.


From the sketch you will see:


α_g = β/2


α_e = 90° - β/2





You get the components of the resultant force by adding the components of the gravity force and electric force:


F_rs = F_g·cos(β/2) + F_e·cos(90°-β/2)


= F_g·cos(β/2) + F_e·sin(β/2)


= 0.987N


F_rp = F_g·sin(β/2) + F_e·sin(90°-β/2)


= F_g·sin(β/2) + F_e·cos(β/2)


= 0.987N = 0.193N





(c)


Sketch sketch a ball with the three forces (as vectors)


resulting force Fr, tension T and inert force m·a (which is a fictitious force).


The balance of these forces leads to the equation of motion:


F_r + T - m·a = 0 → m·a = F_r + T





On the on hand the tension is a force which can only along the direction of the string. On the other hand the acceleration vector can shows in the direction of motion. Since the ball can only move like a pendulum, this direction is perpendicular to the string. So if you break the vector equation above into components with respect to the string direction you will get:


T = -F_rs


m·a = F_rp





The tension in the string:


T = -F_rs = -0.987N


The negative sign indictaes the force acting in diffrent direction.








(d)


m·a = F_rp


→


a = F_rp / m = 0.193N / 0.1kg = 1.93m/s²
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