
A charged cork ball having mass \(1 g\) and charge \(q\) is suspended on a light string in a uniform…

- \(q=11 \times 10^{-8} \mathrm{C}\)
- \(T=5.55 \times 10^{-3} \mathrm{~N}\)
- \(q=12 \times 10^{-9} \mathrm{C}\)
- \(T=4.55 \times 10^{-3} \mathrm{~N}\)
Solution

Force on cork ball due to electric field \(\mathbf{E}\) \(\mathbf{F}=q \mathbf{E}\) According to given diagram, resolving all forces in two given direction, \(\begin{array}{ll} & T \sin \theta=q E_x \quad \ldots (i) \\ & T \cos \theta+q E_y=m g \\ & T \cos \theta=m g-q E_y \quad \ldots (ii) \end{array}\) From Eqs. (i) and (ii), we have \(\frac{T \sin \theta}{T \cos \theta}=\frac{q E_x}{m g-q E_y}\)

\(\begin{aligned} & \tan \theta=\frac{q E_x}{m g-q E_y} \\ & \Rightarrow \quad \tan 37^{\circ}=\frac{q \times 3 \times 10^5}{10^{-3} \times 10-q \times 5 \times 10^5} \\ & \frac{3}{4}=\frac{3 q \times 10^5}{10^{-2}-5 q \times 10^5} \\ & \Rightarrow \quad 3 \times 10^{-2}-15 q \times 10^5=4 \times 3 q \times 10^5 \\ & 0.03-15 \times 10^5 q=12 \times 10^5 q \Rightarrow q=11 \times 10^{-8} \mathrm{C} \end{aligned}\) From Eq. (i), we get \(\begin{aligned} & T \sin 37^{\circ}=1.1 \times 10^{-8} \times 3 \times 10^5 \\ & T \times 0.6=3.3 \times 10^{-3} \\ & T=\frac{3.3}{0.6} \times 10^{-3}=5.5 \times 10^{-3} \mathrm{~N} \end{aligned}\)
Asked in: AP EAMCET 2020 (21 Sep Shift 1)