Two hard spheres, each of mass m = 15.1 g, are rubbed with fur on a dry and are

Question

Two hard spheres, each of mass m = 15.1 g, are rubbed with fur on a dry and are then suspended with insulating strings of length L = 5.45 cm whose support points are a distance d= 2.97 cm from each as shown in the figure below. During the rubbing process, one sphere receives exactly twice the charge of the other. They are observed to hang at equilibrium, each at a angle of theta = 10.9 degree with the vertical. Find the amount of charge on each sphere. (Enter your answers from smallest to largest.)

Explanation / Answer

Charge on the sphere = q

The following three forces acting on the two spheres

1. its weight mg

2. tension T in the string

3. coulomb force

As the forces are in equilibrium, we can write

Let us assume that the two spheres making an angle with the vertical line is forming a triangle, join the inclined and vertical line horizontally, then

[F / AB] = [mg / OA] = [T / OA]

F = mgX[AB/OA] -------(1)

Coulonb force between the two spheres

F = [9X109XqX2q] / r2 --------(2)

9X109 - Constant value

r - Distance between the two spheres

sin(theeta) = opposite/hyp. = x/L ------(3)

theeta = 10.90

L - Length of the string [L = 5.45 cm = 0.0545 m]

sin(10.90) = 0.189

eq. 3 becomes

0.189 = x/0.0545

x = 0.189X0.0545 = 0.01 m

So the distance between the two spheres r = x + d + x

d = 2.97 cm = 0.0297 m [given data]

r = 2x+d = (2X0.01) + 0.0297 = 0.0497 m

Distance between the two spheres r = 0.0497 m

equate eq.(1) and eq.(2), we get

[9X109X2q2] / r2 = mgX[AB/OA] -------(4)

AB = x = 0.01 m

cos(theets) = Adjacent/hyp. = OA/L

cos(10.90) = 0.982, L = 5.45 cm = 0.0545 m

OA = 0.0545X0.982 = 0.0535 m

[9X109X2q2] / r2 = mgX[AB/OA] -------(4)

m - mass of the sphere

g - acceleration due to gravity [g = 9.8 m/s2]

[9X109X2q2] / (0.0497)2 = 0.0151X9.8X[0.01/0.0535]

r = 0.0497 m

m = 15.1 g = 0.0151 Kg

g = 9.8 m/s2

AB = 0.01 m

OA = 0.0535 m

[9X109X2q2] / (0.0497)2 = 0.0151X9.8X[0.01/0.0535]

[ (18X109Xq2) / (2.47009X10-3) ] = 0.02766

7.28718X1012Xq2 = 0.02766

q2 = 3.79568X10-15 C

q2 = 37.9568X10-16 C

q = 6.161X10-8 C [Charge on the one sphere]

2q = 12.3218X10-8 C [Charge on the other sphere]

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