Question

1. How much work does the electric field do in moving a -8.6

Answer 1

1) W=-qV(ba)=-(-8.6x10^-6)65 = 5.59 x 10^-4 Joules

2) its given that the energy acquired by the electron is 3.04 x 10^-16 joules

then the potential difference would be (3.04 x 10^-16)/1.6 x 10^-19
= 1.9 x 10^ 3 Volt

3) E = V/d

= 220 / (3.8 x10^-3)
=57.9 x 10^ 3 N/C

4) voltage = electric field strength*separation distance

V = Ed = 650*0.009= 5.85 V

5) 155/3000000 metres = 5.16 x 10^5 m

6)Potential at a distance r from a point charge q is given by

V = (1/4pi epsilon)(q/r)
where epsilon is the permittivity of the medium in which the charge is placed. Let the medium be air or vacuum.
Then, epsilon = 8.854 x 10^-12 C^2N^-1m^-2
or 1/(4pi epsilon) = 9 x 10^9 Nm^2C-2
Given, q = 3.2 microCoulomb =3.20 x 10^-6 C
r = 10.5 cm = 10.5 x 10^-2 m
V = 9 x 10^9 x 3.2 x 10^-6/(10.5 x 10^-2)
= 2.74 x 10^5 V

7 ) q = (V * r) / k_e (k_e is the electric constant = 8.99*10^9)

= (106 * 0.28) / 8.99*10^9
= 3.3 * 10^-9 C

=3.3nC

8 ) E = (1/2)CV^2

E = (1/2)(8000e-9)(510)^2
E = 1.04 J