Answer
(a) The left plate has the greater potential
(b) $1.18\times 10^3V$
Work Step by Step
(a) We know that the charge moves in the direction of electric field and as the field always points from high potential to low potential, the plate on the left must be at higher potential.
(b) The required potential difference can be determined as follows:
$\Sigma F_x=0$
$\implies Tsin\theta-qF=0$
$E=\frac{Tsin\theta}{q}$.......eq(1)
Similarly $\Sigma F_y=0$
$Tcos\theta-mg=0$
$T=\frac{mg}{cos\theta}$.....eq(2)
Substituting the values of $T$ from eq(2) in eq(1), we obtain:
$E=\frac{mgsin\theta}{cos\theta}.\frac{sin\theta}{q}$
$E=\frac{mg}{q}.tan\theta$
Now the potential difference is given as
$\Delta V=Ed$
$\implies \Delta V=(\frac{mg}{q}tan\theta)d$
We plug in the known values to obtain:
$\Delta V=frac{(0.081Kg)(9.8m/s^2)(tan22^{\circ})(0.025m)}{6.77\times 10^{-6}C}$
$\Delta V=1.18\times 10^3V$
