Question

The distance between two horizontal parallel plates
is 2.0 cm and the potential difference between them
is 120 V. Calculate : (i) electric field between the
plates, (ii) force on an electron passing in between
the plates, (iii) increase in electrical energy of the
electron in moving from one plate to the other.

Answer 1

(i) electric field = electric potential/ seperation between plates

given, electric potential = 120 V

and seperation between plates = 2cm

so, electric field = 120V/2cm =120V/(2 × 10^-2 m) = 6000 N/C

(ii) F = eE , where e is charge on electron

so, e = 1.6 × 10^-19 C and E = 6000N/C

now F = 1.6 × 10^-19 × 6000

= 9.6 × 10^-16 N

(iii) increase in electrical energy of the electron in moving from one plate to the other , U = eV

= 1.6 × 10^-19 C × 120 volts

= 1.92 × 10^-17 J