(c22p7) Two identical conducting spheres, fixed in place, attract each other wit

Question

(c22p7) Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.7767 N when separated by 50 cm, center-to-center. 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 0.2525 N. What were the initial charges on the spheres? Since one is negative and you cannot tell which is positive or negative, there are two solutions. Take the absolute value of the charges and enter the smaller value here. Submit Answer Tries 0/99 Enter the larger value here. Tries n

Explanation / Answer

Let the charge on two identical spheres is q

separated by 0.5 m

having the force (attractive ) between them is Fa = -0.7767 N

we know that the force between two opposite charges is an attractive force whic is a -ve and

the force between the two same charges the force is repulsive force +ve force

that is F = k*q1*q2/r^2

F-ve ===> attractive force

F +ve ==> repulsive force

so given  

Fa = -0.7767 N

kq^2 /r^2 = 0.7767

substituting the values  

9*10^9*q^2/(0.5)^2 = 0.7767

solving for q value  

we have two values q = 4.645*10^-6 C and q = -4.645*10^-6 C

when these spheres are connected with a thin conducting wire so that they were repelling each other

means both attain the same charge that might be -ve charge or +ve charge

so  

F = kq^2/r^2

0.2525 = 9*10^9*q^2/(0.5)^2

q = 2.6484*10^-6 C

or

q = -2.6484*10^-6 C

the initial charge is q = ((4.645*10^-6)+(2.6484*10^-6))/2 C = 7.2934*10^-6/2 C = 3.6467*10^-6 C

or  

q = ((4.645*10^-6)-(2.6484*10^-6) )/2C = 9.983*10^-7 C

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