1
Question
The plates of a capacitor are separated by 5 mm and they are connected to a battery of 100 volt what will be the force on an electron located between the plates?
The plates of a capacitor are separated by 5 mm and they are connected to a battery of 100 volt what will be the force on an electron located between the plates?
Open in App
Solution
Electric field E = V/d, where V is potential difference, and d is distance between the two plates.
so, here E= 100 V/5mm = 100/(5×10^-3)=20x10^3=2×10^4 (Volt/m)
So force on a electron in electric field is F= E.q = (2×10^4)x(1.6x10^-19) Newton
=3.2×10^-15 Newton
Electric field E = V/d, where V is potential difference, and d is distance between the two plates.
so, here E= 100 V/5mm = 100/(5×10^-3)=20x10^3=2×10^4 (Volt/m)
So force on a electron in electric field is F= E.q = (2×10^4)x(1.6x10^-19) Newton
=3.2×10^-15 Newton
Suggest Corrections
2
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
