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# CBSE Class 12-science Answered

electric charges
Asked by rameshsanju123 | 08 Feb, 2024, 08:45: PM
Expert Answer

Figure shows the forces acting on pith ball when it is in equilibrium.

Tension force T acting on the rope is resolved into Tcosθ and Tsinθ along

vertical and horizontal directions respectively.

Where θ is the angle made by the string with vertical as shown in figure

Vertical component of tension force balances the weight of the ball .

T cosθ = m g

where m = 0.5 kg is mass of the ball and g is acceleration due to gravity .

Above expression is written as

T cosθ = 0.5 × 9.8  N = 4.9 N ............................(1)

Horizontal component of tension force balances the electrostatic force Fe between charged balls.

T sinθ = Fe = K × ( q2 / d2 )

where K = 1/(4πεo ) = 9 × 109 N m2 C-2 is Coulomb constant,  q = 5 μC is the charge on each pith ball

and d is distance between pith balls .

Hence above expression is written as

..........................(2)

From (1) and (2) , we get

.......................(3)

when θ is small , tanθ ≈ θ = (arc)/radius = (d/2) / 1 = d/2

Hence eqn.(3) becomes

d3 = 0.092 m3

hence , we get d = 0.451 m

Answered by Thiyagarajan K | 09 Feb, 2024, 08:31: AM

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