A pendulum bob of mass 80 mg and carrying a charge of 2 - 10-8 C is at rest in a uniform horizontal electric field of 20 kV m -1 Find the tension in the thread-

From the figure, T cosθ = mg T sinθ = qE tanθ = qE/mg = 20 × 10^3 × 2 × 10^ 8/80 × 10^ 6× 9.8 tanθ = 40/8 × 9.8 = 0.51 θ = 27^∘ From equation (i) , T = m ...

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Standard XII

Physics

Tension in a String

A pendulum bo...

Question

A pendulum bob of mass 80 mg and carrying a charge of $2 \times 10^{- 8} C$ is at rest in a uniform horizontal electric field of $20 k V m^{- 1}$ Find the tension in the thread.

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Solution

From the figure,
$T cos θ = m g T sin θ = q E tan θ = \frac{q E}{m g} = \frac{20 \times 10^{3} \times 2 \times 10^{- 8}}{80 \times 10^{- 6} \times 9.8} tan θ = \frac{40}{8 \times 9.8} = 0.51 θ = 27^{\circ}$

From equation (i) ,
$T = \frac{m g}{cos θ} = \frac{80 \times 10^{- 6} \times 9.8}{cos 27^{\circ}} T = 8.8 \times 10^{- 4} N$

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A pendulum bob of mass 80 mg and carrying a charge of 2×10−8C is at rest in a uniform horizontal electric field of 20kVm−1 Find the tension in the thread.

From the figure, Tcosθ=mgTsinθ=qEtanθ=qEmg=20×103×2×10−880×10−6×9.8tanθ=408×9.8=0.51θ=27∘ From equation (i) , T=mgcosθ=80×10−6×9.8cos27∘T=8.8×10−4N

A pendulum bob of mass 80 mg and carrying a charge of $2 \times 10^{- 8} C$ is at rest in a uniform horizontal electric field of $20 k V m^{- 1}$ Find the tension in the thread.