Two 13.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14 V battery. What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process. What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process. What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.7 cm apart? The electrodes remain connected to the battery during this process.

Explanation:

The capacitor is of parallel plate capacitor type

Capacitance C = ε₀ A / 4π d

ε₀  is 8.85 x 10⁻¹² , A  is plate area and  d is distance between plate .

d = 1.7 cm

C = 8.85 x 10⁻¹² x π x (6.5 x 10⁻² )² / 4π x 1.7 x 10⁻²

= 55 x 10⁻¹⁴F .

Charge on each electrode = C x V , V is voltage of battery .

= 55 x 10⁻¹⁴ x 14

= 770 x 10⁻¹⁴ C  

Electric field strength = V / d where V is potential difference of battery , d is distance between plate .

= 14 / 1.7 x 10⁻²

= 8.23 x 10² V / m

The potential difference between plate

= potential difference of the battery

= 14 V .

t.                 w               e                 n                   t                 y                 c             h               a           r           a               c                 t             e             r             s        

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer