Answer:
a
[tex]V = -47.65 N/C[/tex]
b
At the origin
Explanation:
From the question we are told that
The initial speed is [tex]v_1 = 4.10 *10^{6} \ m/s[/tex]
The speed at (x = 2.00 cm) is [tex]v_f = 1.76 *10^{5} \ m/s[/tex]
Generally the electric potential difference is mathematically represented as
[tex]V = \frac{W}{q}[/tex]
Here W is the work-done which is mathematically represented as
[tex]W = K_f - K_i[/tex]
Here [tex]K_f[/tex] is the kinetic energy at x = 2.00 cm mathematically expressed as
[tex]K_f = \frac{1}{2} * m* v^2_f[/tex]
and
[tex]K_i[/tex] is the kinetic energy at origin mathematically expressed as
[tex]K_f = \frac{1}{2} * m* v^2_i[/tex]
So
[tex]V = \frac{1}{q} [ \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2][/tex]
[tex]V = \frac{m}{2q} [v_f^2 - v_i^2][/tex]
Here m is the mass of electron with value [tex]m = 9.1*10^{-31} \ kg[/tex]
q is the charge on the electron with value [tex]q = 1.60*10^{-19} \ C[/tex]
So
[tex]V = \frac{(9.1 *10^{-31})}{2(1.60*10^{-19})} [(1.76*10^{5})^2 - (4.10*10^{6})^2][/tex]
[tex]V = -47.65 N/C[/tex]
So given that the difference is negative then potential is higher at the origin
