(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 -0.3826 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 0.1231 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.

Explanation / Answer

lets say the initial charges were q1 and -q2

0.3826= k q1 q2/0.5^2

q1= 0.0106*10^-9/q2

then becuase of conducting wire the charges will be shared equally i.e q1-q2/2

so

0.1231 = k(q1-q2)^2/4 *0.5^2

q1-q2 = 0.03698 * 10^-4

0.0106*10^-9/q2 - q2= 0.03698*10^-4

q2=x

0.0106*10^-9-x^2=0.03698*10^-4*x

x^2+0.03698*10^-4x-0.0106*10^-9=0

Qudratic equation a=1,b=0.03698*10^-4, c= -0.0106*10^-9

x= -b +- Sqrt(b^2- 4ac) / 2a

-0.03698*10^-4 +- Sqrt((0.03698*10^-4)^2- 4*1*-0.0106*10^-9))/2

(-0.03698*10^-4 +-0.000007488)/2

1.895*10^-6 C , -5.59*10^-6 C

Check the calculations once , Also Enter the numbers without negative sign

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