Dear student From the free body diagram of the blockTcosθ=mg 1and for horizontal equilibriumTsinθ=14π∈q2(2lsinθ)2=14π∈q24l2sin2θ=q216π∈l2sin2θ 2dividing eqn 2 by 1TsinθTcosθ=q216π∈l2sin2θmgtanθ=q2mg×16π∈l2sin2θtanθ×sin2θ=q2mg×16π∈l2differentiating both sides with respect to time tanθ×d(sin2θ)dt+sin2θ×d(tanθ)dt=q2mg×16π∈l2tanθ×(2sinθ×cosθ)dθdt+sin2θ×sec2θ×dθdt=q2mg×16π∈l2(2sin2θ+tan2θ)dθdt=2qmg×16π∈l2dqdt(2sin2θ+tan2θ)×ω=q8mg×π∈l2dqdtdqdt=(2sin2θ+tan2θ)×ω×8mg×π∈l2qRegards

Question 7?

Question 7? 6. 7. A charge Q is placed at a distance a/2 above the centre of a horizontal, square surface of edge 'a' as shown in the figure, While a second charge Q is placed ata distance 'a' below one of the vertices of the square. Find the a flux of the electric field through the square. a 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 C. The angle between the string is 20, and = (Do, due to leakage of charge at a very slow rate. Find the rate of initially changes at a rate — dt dq leakage of charge (i.e. — dt Ltd., House, 294, Kalu Sarai, sarvapnya Vihar, New Delhi-110016, Ph 26515949, 26569493, Fax 26513942

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F r o m t h e f r e e b o d y d i a g r a m o f t h e b l o c k T \cos θ = m g . . . 1 a n d f o r h o r i z o n t a l e q u i l i b r i u m T \sin θ = \frac{1}{4 π \in} \frac{q^{2}}{(2 l \sin θ)^{2}} = \frac{1}{4 π \in} \frac{q^{2}}{4 l^{2} \sin^{2} θ} = \frac{q^{2}}{16 π \in l^{2} \sin^{2} θ} . . . 2 d i v i d i n g e q n 2 b y 1 \frac{T \sin θ}{T \cos θ} = \frac{\frac{q^{2}}{16 π \in l^{2} \sin^{2} θ}}{m g} \tan θ = \frac{q^{2}}{m g \times 16 π \in l^{2} \sin^{2} θ} \tan θ \times \sin^{2} θ = \frac{q^{2}}{m g \times 16 π \in l^{2}} d i f f e r e n t i a t i n g b o t h s i d e s w i t h r e s p e c t t o t i m e \tan θ \times \frac{d (\sin^{2} θ)}{d t} + \sin^{2} θ \times \frac{d (\tan θ)}{d t} = \frac{q^{2}}{m g \times 16 π \in l^{2}} \tan θ \times (2 \sin θ \times \cos θ) \frac{d θ}{d t} + \sin^{2} θ \times s e c^{2} θ \times \frac{d θ}{d t} = \frac{q^{2}}{m g \times 16 π \in l^{2}} (2 \sin^{2} θ + \tan^{2} θ) \frac{d θ}{d t} = \frac{2 q}{m g \times 16 π \in l^{2}} \frac{d q}{d t} (2 \sin^{2} θ + \tan^{2} θ) \times ω = \frac{q}{8 m g \times π \in l^{2}} \frac{d q}{d t} \frac{d q}{d t} = \frac{(2 \sin^{2} θ + \tan^{2} θ) \times ω \times 8 m g \times π \in l^{2}}{q} R e g a r d s

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