(8c21p7) Two identical conducting spheres, fixed in place, attract each other wi

Question

(8c21p7) Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.6559 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 0.0562 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. Enter the larger value here.

Explanation / Answer

By using the Coulomb's law of electrostatics the force between two satationary charges q1,q2, separated by a distance r then the electrostatic force between them is

F= k*q1*q2/r^2

now given

F = kQq/r²

-0.6559 N = (9 * 10^9)Qq/0.5²

Qq = -18.219 * 10^(-12) C^2

But since we are told that the charges attract one another, we know that q1 and q2 have
opposite signs and so their product must be negative.

Qq = -18.219* 10^(-12) C^2

Q = -18.219 * 10^(-12)/q. . . . . . . . . . . . . . . (1)

Then the two spheres are joined by a wire, If the new charge on each sphere is Q',

Q' + Q' = Q + q = 2Q'

F = kQ'q/r²

0.0562 N = (9 * 10^9)(Q')²/0.5²

Q'^2 = 1.561*10^-12 C^2

Q'= 1.24944*10^-6 C= 1.25*10^-6 C

Q + q = 2Q'

Q + q = 2 * 1.25*10^-6 C. . . . . . . . . . . . . . . (2)

substitute equation 1 to 2,

(-18.219* 10^(-12)/q )+ q = 2 * 1.25*10^-6

q = 5.6966*10^-6 C

or

q = -3.1966*10^-6 Coulomb

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.