Two identical conducting spheres, fixed in place, attract each other with an ele

Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.2661 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 0.3682 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

let the initial charge on spheres be Q1 and Q2

r = distance between the spheres = 0.50 m

Fattaract = force of attaraction between the spheres = k Q1 Q2/r2

- 0.2661 = (9 x 109) Q1 Q2 /(0.5)2

Q1 Q2 = - 7.4 x 10-12

Q1 = - 7.4 x 10-12 /Q2                      eq-1

after connecting the charge by conducting wire , they have equal charge which is given as

Q = (Q1 + Q2 )/2                 eq-2

Frepel = force of repelsion = 0.3682 = k Q2/r2

0.3682 = (9 x 109) Q2/(0.5)2

Q = 3.2 x 10-6      or - 3.2 x 10-6

using eq-2

Q1 + Q2 = 2 Q = 2 (3.2 x 10-6 ) = 6.4 x 10-6

using eq-1

(- 7.4 x 10-12 /Q2 ) + Q2 = 6.4 x 10-6

Q2 = - 1 x 10-6

Q1 = 7.4 x 10-6

or

Q1 + Q2 = - 6.4 x 10-6

using eq-1

(- 7.4 x 10-12 /Q2 ) + Q2 = - 6.4 x 10-6

Q2 = - 7.4 x 10-6

and

Q1 = - 1 x 10-6

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