Question

g Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.7214 N when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0562 N. What were the initial charges on the spheres?

Answer 1

Answer:

q_2 = (+,-) 5.9 *10^-6 C , q_2 = (+,-) 3.4*10^-6 C

q_1 = (+,-) 3.4*10^-6 C    , q_1 = (+,-) 5.9*10^-6 C

q_3 = q_4 = (+, -) 1.25*10^-6 C

Explanation:

Given:

F_1,2 = 0.7214 N

F_3,4 = 0.0562 N

separation r = 0.5 m (remains constant)

q_1  and q_2 opposite charged before transfer

q_3 and q_4 same charge after transfer

Solution:

F_1,2 = k*q_1*q_2 / r^2 = -0.7214 N (Attraction)  

q_1*q_2 = -2.00612*10^-11 .... 1

F_3,4 = k*q_3*q_4 / r^2 = +0.0562 N (Repulsion)  

q_3*q_4 = 1.5628*10^-12 .... 2

Conservation of charge:

Since, no charge is lost the total charge before and after transfer must be same. Hence,

q_1 + q_2 = q_3 + q_4 ..... 3

Also, electrons are transferred till the all the positive charge is neutralized. Such that after transfer both charges are equal, Hence:

q_3 = q_4   ..... 4

Now we have 4 equations and four unknowns q_1, q_2, q_3, and q_4. These equations can now be solved simultaneously as follows:

Solving for unknowns:

Using 3 and 4

q_3 = q_4 = 0.5*( q_1 + q_2 )    .... 5

Using 1 , 2, and 5

q_1 = -2.00612*10^-11 / q_2   ..... 6

0.25*( q_1 + q_2 )^2 = 1.5628*10^-12 .... 7

Using 6 and 7

(q_1 ^2 + 2*q_1*q_2 + q_2^2) = 6.2512*10^-12

(4.025*10^-22 / q_2^2) - (4.01224*10^-11) + q_2^2) =  6.2512*10^-12

q_2^4 - q_2^2 *4.6374*10^-11 + 4.025*10^-22 = 0

Making substitution for disguised quadratic:

q_2^2 = x

x^2 - x^2 *4.6374*10^-11 + 4.025*10^-22 = 0

$x_1 = \frac{(4.6374*10^-11)+\sqrt{(4.6374*10^-11)^2 - 4(1)*(4.025*10^-22)} }{2} \\\\x_1 = \frac{(4.6374*10^-11)+ (2.325*10^-11)}{2} \\\\x_1 = 3.4812*10^-11\\\\x_2 = \frac{(4.6374*10^-11)-\sqrt{(4.6374*10^-11)^2 - 4(1)*(4.025*10^-22)} }{2} \\\\x_2 = \frac{(4.6374*10^-11)- (2.325*10^-11)}{2} \\\\x_2 = 1.15622*10^-11$

Hence,

After Back substitution:

q_2 = (+,-) 5.9 *10^-6 C , q_2 = (+,-) 3.4*10^-6 C

q_1 = (+,-) 3.4*10^-6 C    , q_1 = (+,-) 5.9*10^-6 C

q_3 = q_4 = (+, -) 1.25*10^-6 C