Two metal spheres, each of mass 0.3kg, are positively charged by an electrostati
ID: 1449223 • Letter: T
Question
Two metal spheres, each of mass 0.3kg, are positively charged by an electrostatic generator. Each is attached to an identical insulating string and suspended from the same point. Sphere A has three times the charge of sphere B. They repel eachother and hang with separation 0.5m. The length of the string from the point of support to the center of the ball is 2.5m.
1) Draw the electric field lines for this arrangement.
2) What is the electrostatic force on each sphere?
3) What is the charge on each sphere?
4) The two spheres are briefly connected by a copper wire. What is the final charge on each sphere?
5) What now is the separation distance between the two spheres?
Please help by showing work.
Explanation / Answer
angle made by string with string is @ then
sin@ = (0.5/2) / 2.5
@ = 5.74 deg
suppose tension in string is T then
in in vertical direction,
Tcos@ = mg
T cos5.74 = 0.3 x 9.8
T = 2.955 N
in horizontal direction,
Tsin5.73 - Fe = 0
Fe = 2.955 sin5.74 = 0.30 N ..........Ans (2)
and electric force, Fe = k (q1)(q2) / d^2
qA = 3 qB
Fe = k (3qB) (qB) / d^2
0.30 = (9 x 10^9 x 3 x qB^2 ) / (0.5^2)
qB = 1.65 x 10^-6 C
so charge pn sphere B = 1.65 x 10^-6C
charge on sphere A = 3qB = 4.96 x 10^-6 C ......Ans(3)
4) when connected by wire , total charge will half -half in both sphere .
total charge = qA + qB = 6.617 x 10^-6 C
now charge on qA = qB = 3.308 x 10^-6 C
Tcos@ = mg
Tsin@ = Fe = (9 x 10^9 x (3.308 x 10^-6)^2) / d^2 )
and d = 2L sin@ = 5 sin@
T sin@ = (3.94 x 10^-3) / sin^2 @
putting T from 1st equation,
(0.3 x 9.8 / cos@) sin^3 @ = 3.94 x 10^-3
sin^3 @ / cos@ = 1.34 x 10^-3
sin^3 / sqrt(1 - sin^2@) = 1.34 x 10^-3
sin^6 @ / 1 - sin^2@ = 1.796 x 10^-6
556712.98 sin^6 @ = 1 - sin^2 @
sin^6 @ + sin^2@ - 1 = 0
sin^2@ = 0.012155
@ = 6.33 deg
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