Question

A point charge q = ?2.3 nC is initially at rest adjacent to the negative plate of a capacitor. The charge per unit area on the plates is 4.3 C/m² and the space between the plates is 6.2 mm.

(a) What is the potential difference between the plates?
kV

(b) What is the kinetic energy of the point charge just before it hits the positive plate, assuming no other forces act on it?

Answer 1

a) E = sigma/epsilon)

= 4.3*10*10^-6/8.854*10^-12

= 4.857*10^5 N/c

V = E*d
= 4.857*10^5*6.2*10^-3
= 3.01*10^3 volts
= 3.01 kVolts <<<<<<<<<<<<--------answer

b) Workdone = q*deltaV

= 2.3*10^-9*3.01*10^3

= 6.923*10^-6J

= 6.923 micro J

kinetic energy = workdone = 6.923 micro J <<<<<<<<---------answer

Answer 2

a) Potential defference = 4.3*6.2*10^-3/ (8.85*10^-12) =3012425 k V

b) kinetic energy = 2.3*10^-9 *3012.42*10^6 =6.92 J

Answer 3

a) Q =C V

C = epsilon A/d = 1.43*10^-9

V = 3.011*10^9 V

b) K.E = qV = 6.29 J

Answer 4

1) V= E*d= sigma/epsilon*d = 4.3*0.0062/8.85*10^-12

=0.003012*10^12V

2)change in KE= work done = energy stored in capacitor= 1/2*q*V= 0.5*(-2.3*10^-9*0.003012*10^12)

=3.4638Nm

Answer 5

sigma = Q/A

= 4.3*10*10^-6/8.854*10^-12= 4.857*10^5 N/c

V = E*d

= 4.857*10^5*6.2*10^-3

= 3.01*10^3 volts

= 3.01 kVolts

potential difference = Ed

potential difference = 4.3*6.2*10^-3/ (8.85*10^-12) =3012425 k V

b) Workdone = q*deltaV

= 2.3*10^-9*3.01*10^3

= 6.923*10^-6J

= 6.923 micro J

kinetic energy = workdone = 6.923 micro j