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202301110923 Solution to 1980-AL-PHY-IA-26

A charged ball X is suspended by a string.

When a uniform electric field E is applied horizontally, the ball is displaced a horizontal distance a such that the string makes an angle \theta with the vertical.

Roughwork.

As always start with a free-body diagram. Hence, draw

Resolving components, write

\begin{cases} \textrm{Horizontal:}\quad & T\sin\theta = qE \\ \textrm{Vertical:}\quad & T\cos\theta = mg \\ \end{cases}

Squaring and summing, write

\begin{aligned} (T\sin\theta )^2+ (T\cos\theta)^2 &= (qE)^2 + (mg)^2\\ T & = \sqrt{q^2E^2+m^2g^2} \\ \end{aligned}

By considering the electric field strength E as an independent variable, and the angle of elevation \theta the dependent variable, make differentiation

\displaystyle{\mathrm{d}E = \bigg(\frac{T\cos\theta}{q}\bigg)\,\mathrm{d}\theta}\qquad(\theta\in [0,90^\circ])

This problem is not to be attempted.