Two pith balls carrying equal charges are suspended from a comman point by strings of equal length , the equililibrium seperation between them is r. now the strings are rigidly clamped at half the hieght The equililibrium seperation between balls now become

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Q. Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is

r

. Now the strings are rigidly clamped at half the height. The equilibrium seperation between the balls now become

Q. Two pith balls carrying equal charges are suspended from a common point by strings of equal length, the equilibrium separation between them is

r

. Fig. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls, now becomes.

Q. Two pith balls carrying identical charges are suspended from a common point by strings of equal lengths, the equilibrium separation between them is

r

. Now the strings are rigidly clamped at half the height. The equilibrium separation between the balls now will be :

Two identical pith balls, each carrying a charge q, are suspended from a common point by two strings of equal length l. Find the mass of each ball if the angle between the strings is $2 θ$ in equilibrium.

Q. Two identical pith balls, each of mass

m

and carrying identical charges

q

each, are suspended from a common point by two strings of equal length

l

. The angle between the string is

2 θ

, and initially changes at a rage

∣ ∣ ∣ \frac{d θ}{d t} ∣ ∣ ∣ = ω_{0}

, due to leakage of charge at a very slow rate. Find the rate of leakage of charge (i.e.

∣ ∣ ∣ \frac{d q}{d t} ∣ ∣ ∣

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