Question

Your friend sets up a uniform electric field (in units of N/C):

E (x,y)=−1400i^−4800j^

a. What do you predict you will measure to be the voltage difference for the displacement going from position (-26,-25) to position (-28,-45) with distances measured in centimeters?

b. Your friend resets the electric field. This time you need to figure out the components of the field yourself. You measure the voltage difference along the positive xx direction to be 2000 V across a distance of 4 cm. The voltage difference along the positive y direction is -3800 V across a distance of 2 cm. [Enter your results for Ex in the first box and Ey in the second.]

c. A proton is placed in the electric field and starting from rest, it is allowed to freely travel a distance of 5 cm before it is recaptured. What is the voltage difference it crosses during its motion? [NOTE: Use the field data you computed in the previous question.]  

d. How fast is the proton moving just before it is recaptured?

Answer 1

(a)

Given that,

E = -1400 i - 4800 j

dr = (-28 i - 45 j) - (-26 i - 25 j) = (-2 i - 20 j)cm

dr = (-0.02 i - 0.20 j) m

We know that,

E = -dV / dr

dV = -E.dr

dV = ( -1400 i - 4800 j) .  (-0.02 i - 0.20 j)

dV = 28 + 960

dV = 988 V

(b)

E = -dVx / x - dVy / y

E = -(2000 / 0.04)i - (-3800 / 0.02)j

E = (-50000 i + 190000 j) V/m

(c)

Magnitude of electric field,

E = sqrt (50000^2 + 190000^2) = 196468.82 N/C

Potential difference, V = E.d

V = 196468.82 . 0.05

V = 9823.44 V

(d)

(1/2)mv^2 = qV

(1/2)*1.67*10^(-27)*v^2 = 1.6*10^(-19)*9823.44

v = 1.37*10^6 m/s