A small sphere of mass $20 g$ and having charge $0.5 m C$ is suspended by a thread between the plates of a parallel plate capacitor as shown in the figure. What potential difference ( in volt) between the plates (separation $20 c m$ ) should be applied so that the thread makes an angle of $45^{0}$ with the vertical?

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Solution

Here, $T c o s 45 = m g . . . . . . (1)$
$T s i n 45 = q E . . . . . . . (2)$
$(2) / (1), t a n 45 = 1 = \frac{q E}{m g}$
Hence electric field, $E = \frac{m g}{q}$
$∴$ P.D. between the plates, $V = E d = \frac{m g d}{q}$
or $V = \frac{20 \times 10^{- 3} \times 10 \times 20 \times 10^{- 2}}{0.5 \times 10^{- 3}} = \frac{40}{0.5} = 80 V$

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Here, Tcos45=mg......(1) Tsin45=qE.......(2)(2)/(1),tan45=1=qEmgHence electric field, E=mgq∴P.D. between the plates, V=Ed=mgdqor V=20×10−3×10×20×10−20.5×10−3=400.5=80V