Question

The voltage difference between two charged parallel plate conductors separated by a distance of 14 mm is 3 V. (a) What is the magnitude of the electric field between the plates? b) What direction does the electric field go? (c) What force would this electric field apply to an electron? (d) What force would this electric field apply to a proton? (e) How much work would be done on/by an electron that moves from the negatively charged plate to the positively charged plate? (f) How much work would be done on/by an electron that moves from the positively charged plate to the negatively charged plate?

Answer 1

Answer:

Explanation:

a ) Electric field = Potential difference between plates / distance between them

= $\frac{3}{14\times 10^{-3}}$

2142 .85 V/m

b ) Electric field will go from high volt to low volt.

c ) Force = charge x electric field

= 1.6 x 10⁻¹⁹ x 2142.85 = 3.428 x 10⁻¹⁶ N

d ) Same force will apply on proton since both proton and electron have same charge , only the direction of force will be opposite.

e ) Work done is equal to 2142.85 eV.

f ) Same .